CONIC SECTIONS

THE mathematicians of antiquity regarded the straight line and the circle as the most simple of all geometrical lines, and the celebrated geometer Euclid has employed no other in his well known Elements. By these alone the ancients resolved a great number of problems, of which the more simple are contained in Euclid's Elements; but many of higher difficulty were resolved in his other writings, and in treatises of Archimedes and Apollonius, which have only in part reached our times. There is, however, from the very nature of geometrical science, a limit to the applicability of the straight line and circle. Some problems admit of only one solution: these can be resolved by the intersection of two straight lines. Others again admit of two solutions, and such require lines which intersect each other in two points; therefore they may be resolved by the straight line and circle, or two circles.

If, however, a problem be of such a nature as in its most general form to admit of three solutions, it must necessarily be determined by the intersection of two lines which intersect each other in three points: it therefore cannot be resolved by the straight line and circle alone. Now the ancients had actually proposed to themselves such problems, and in this way it may be supposed they had discovered the necessity of admitting other lines, in addition to the straight line and circle, into their geometry.

The interest which mankind take in mathematical speculations is but little in comparison to that which is excited by works of poetry, oratory, or history; hence it has happened that ancient treatises on these subjects have had a better chance of descending to our times. It is not, therefore, wonderful that none of the works of the more early Greek geometers have reached us, nor have we any work of great antiquity professedly written on the subject of the conic sections. Our curiosity must therefore rest satisfied with the knowledge of a few accidental notices and facts gleaned from different authors.

The discovery of the conic sections seems to have originated in the school of Plato, in which geometry was highly respected and much cultivated. It is probable that the followers of that philosopher were led to the discovery of these curves, and to the investigation of many of their properties, in seeking to resolve the two famous problems of the duplication of the cube, and the trisection of an angle, for which the artifices of the ordinary or plane geometry were insufficient. Two solutions of the former problem, by the help of the conic sections, are preserved by Eutocius,¹ and are attributed by him to Menæchmus, the scholar of Eudoxus, who lived not much posterior to the time of Plato: and this circumstance, added to a few words in an epigram of Eratosthenes,² has been thought sufficient authority, by some authors, to ascribe the honour of the discovery of the conic sections to Menæchmus. We may at least infer that, at this epoch, geometers had made some progress in developing the properties of these curves.

The writings of Archimedes that have reached us explicitly show that the geometers before his time had advanced a great length in investigating the properties of the conic sections. This author expressly mentions many principal propositions to have been demonstrated by preceding writers; and he often refers to properties of the conic sections,

as truths commonly divulged and known to mathematicians. His own discoveries in this branch of science are worthy of the most profound and inventive genius of antiquity. In the quadrature of the parabola he gave the first and the most remarkable instance that has yet been discovered, of the exact equality of a curvilinear to a rectilinear space. He determined the proportion of the elliptic spaces to the circle; and he invented many propositions respecting the mensuration of the solids formed by the revolution of the conic sections about their axes.

It is chiefly from the writings of Apollonius of Perga, a town in Pamphylia, on the subject of the conic sections, that we know how far the ancient mathematicians carried their speculations concerning these curves. Apollonius flourished under Ptolemy Philopator, about forty years later than Archimedes. He formed his taste for geometry, and acquired that superior skill in the science, to which he is indebted for his fame, in the school of Alexandria, under the successors of Euclid. Besides his great work on the conic sections, he was the author of many smaller treatises relating chiefly to the geometrical analysis, the originals of which have all perished, and are only known to modern mathematicians by the account given of them by Pappus of Alexandria, in the seventh book of his Mathematical Collections. (See APOLLONIUS.)

The work of Apollonius on the conic sections, written in eight books, was held in such high estimation by the ancients, as to procure for him the name of the great geometer. The first four books of this treatise only have come down to us in the original Greek. It is the purpose of these four books, as we are informed in the prefatory epistle to Eudemus, to deliver the elements of the science; and in this part of his labour the author claims no further merit than that of having collected, amplified, and reduced to order, the discoveries of preceding mathematicians. One improvement introduced by Apollonius is too remarkable to be passed over without notice. The geometers who preceded him derived each curve from a right cone, which they conceived to be cut by a plane perpendicular to its slant side. It will readily be perceived, from what is shown in the first section of the fourth part of the following treatise, that the section would be a parabola when the vertical angle of the cone was a right angle, an ellipse when it was acute, and a hyperbola when it was obtuse. Thus each curve was derived from a different sort of cone. Apollonius was the first to show that all the curves are produced from any sort of cone, whether right or oblique, according to the different inclinations of the cutting plane. This fact is one remarkable instance of the adherence of the mind to its first conceptions, and of the slowness and difficulty with which it generalizes.

The original of the last four books of the treatise of Apollonius is lost; nor is it easy to ascertain in what age it disappeared. In the year 1658 Borelli discovered at Florence an Arabic manuscript entitled Apollonii Pergai Conicorum Libri Octo. By the liberality of the Duke of Tuscany, he was permitted to carry the manuscript to Rome, and, with the aid of an Arabic scholar, Abraham Ecchellensis, he published in 1661 a Latin translation of it. The manuscript, although from its title it was expected

to be a complete translation of all the eight books, yet was found to contain only the first seven books; and it is remarkable, that another manuscript, brought from the East by Golius, the learned professor of Leyden, so early as 1664, as well as a third, of which Ravius published a translation in 1669, have the same defect. All the three manuscripts agreeing in the want of the eighth book, we may now consider that part of the work of Apollonius as irrecoverably lost. Fortunately, in the Collectiones Mathematicæ of Pappus, in whose time the entire treatise of Apollonius was extant, there is preserved some account of the subjects treated in each book, and all the Lemmata required in the investigations of the propositions they contain. Dr Halley, who in 1710 gave a correct edition of the Conics of Apollonius, guided in his researches by the lights derived from Pappus, has restored the eighth book with so much ability as to leave little room to regret the original.

The last four books of the Conics of Apollonius, containing the higher or more recondite parts of the science, are generally supposed to be the fruit of the author's own researches; and they do much honour to the geometrical skill and invention of the great geometer. Even in our times the whole treatise must be regarded as a very extensive, if not a complete work on the conic sections. Modern mathematicians make important applications of these curves, with which the ancients were unacquainted; and they have been thus led to consider the subject in particular points of view, suited to their purposes; but they have made few discoveries of which there are not some traces to be found in the work of the illustrious ancient.

The geometers who followed Apollonius seem to have contented themselves with the humble task of commenting on his treatise, and of rendering it of more easy access to the bulk of mathematicians. Till about the middle of the 16th century, the history of this branch of mathematical science presents nothing remarkable. The study of it was then revived; and since that time this part of mathematics has been more cultivated, or has been illustrated by a greater variety of ingenious writings.

Among the ancients the study of the conic sections was a subject of pure intellectual speculation. The applications of the properties of these curves in natural philosophy have, in modern times, given to this part of the mathematics a degree of importance that it did not formerly possess. That which, in former times, might be considered as interesting only to the learned theorist and profound mathematician, is now a necessary attainment to him who would not be ignorant of those discoveries in nature that do the greatest honour to the present age.

It is curious to remark the progress of discovery, and the connexion that subsists between the different branches of human knowledge; and it excites some degree of admiration to reflect, that the astronomical discoveries of Kepler, and the sublime theory of Newton, depend on the seemingly barren speculations of Greek geometers concerning the sections of the cone.

Apollonius, and all the writers on conic sections before Dr Wallis, derived the elementary properties of the curves from the nature of the cone. In the second part of his treatise De Sectionibus Conicis, published in 1665, Dr Wallis laid aside the consideration of the cone, deriving the properties of the curves from a description in plano. Since his time authors have been much divided as to the best method of defining those curves, and demonstrating their elementary properties; many of them preferring that of the ancient geometers, while others, and some of great note, have followed the example of Dr Wallis.

In support of the innovation made by Dr Wallis, it is urged, that in the ancient manner of treating the conic

sections, young students are perplexed and discouraged by the previous matter to be learnt respecting the generation and properties of the cone; and that they find it no easy task to conceive steadily, and to understand, diagrams which represent lines drawn in different planes; all these difficulties are avoided by defining the curves in plano from one of their essential properties. It is not our intention particularly to discuss this point; and we have only to add, that in the following treatise we have chosen to deduce the properties of the conic sections from their description in plano, as better adapted to the nature of a work designed for general readers.

A geometrical treatise on the conic sections must necessarily be founded upon the elements of geometry. As Euclid's Elements of Geometry are generally studied, and in every one's hands, we have chosen to refer to it in the demonstrations. The edition we have used is that published by the late Professor Playfair of Edinburgh. Although the references are made to Euclid's Elements, yet they will also apply to the treatise on GEOMETRY given in this work; for a table is there given, indicating the particular proposition of our treatise that corresponds to each of the most material propositions in Euclid's Elements.

The references are to be thus understood: (20, 1, E.) means the 20th prop. of the first book of Euclid's Elements; (2 Cor. 20, 6, E.) means the second corollary to the 20th prop. of the sixth book of the same work; and so of others. Again, (7) means the seventh proposition of that Part of the following treatise in which such reference happens to occur; (Cor. 1) means the corollary to the first proposition; (2 Cor. 3) means the second corollary to the third proposition, &c.—such references being all made to the propositions in the division of the treatise in which they are found.

We shall conclude this brief history of this most interesting part of the ancient geometry, with a catalogue of authors and their works.

Archimedes On the Quadrature of the Parabola, and on Conoids and Spheroids. Archimedes lived about 180 years before the Christian era. The best editions of his works are Torelli's, in Greek and Latin, Oxford, 1792; and Peyrard's French translation, Paris, 1808.

Collectiones Mathematicæ Pappi Alexandrini, lib. vii. Bononiæ, 1660. Here some account of Apollonius' Conics, and the Lemmata, is given. Apollonius lived about forty years later than Archimedes. There have been various editions of his Conics, viz. Apollonii Pergæi Conicorum libri iv. cum Commentariis R. P. Claudii Richardi. Antwerp, 1655.

Apollonii Pergæi Conicorum lib. v. vi. vii., Paraphrase Abalphato Asphalmensi, ex Arabico in Latinum per Abraham Eccellensem Maronitam redditu, cum Notis J. Alfonsi Borelli. Florent. 1661.

Apollonii Pergæi Con. Sect. lib. v. vi. vii. in Græcia deperditu, jam vero ex Arabico M.S. ante quadringentos annos elaborata opera subitanea Latinitate donati a Christiano Rario. Upsal, 1669.

Apollonii Conica Methodo Nova illustrata et succincte demonstrata, per Isaacum Barrow. London, 1675.

Emendatio et Restitutio Conicorum Apollonii Pergæi. Autore Francisco Maurolyco. Colonie, 1675.

Apollonii Pergæi Conicorum libri octo, et Sciri Antissensis de Sectione Cylindri et Coni libri duo. Oxford, 1710. This is Dr Halley's edition. The first four books, together with the Lemmata of Pappus, and the commentaries of Eutocius, have been published from Greek manuscripts, accompanied with a Latin translation; the fifth, sixth, and seventh books, which also contain the Lemmata, have been translated from Arabic into Latin; the eighth

Introduction has been restored by Dr Halley. The books of Serenus are in Greek and Latin.

De Maximis et Minimis, Geometrica Divinatio in quintum Conicorum Apollonii Pergaei, nunc desideratum librum. Autore Vincentio Viviani. Florence, 1659.

De Iocis Solidis, Secunda Divinatio Geometrica in quinque libros injuria temperum omisso Aristaei Senioris Geometra. Aut. Vin. Viviani. &c. Opus Conicorum, continens Elementa Tractatum ejusdem Viviani. Florence, 1701.

Apollonius Catus, a work on Conics in German, by Benjamin Bramer, printed 1634.

Claudii Mydorgii Patricii Parisini Prodroni Catoptrorum et Dioptrorum; sive Conicorum Operis ad abditum radii reflexum et refractum mysteria praevii et faciem praeficientis. Libri iv. priores. Paris, 1641.

Francisci a Schooten Leydensis de Organica Conicorum Sectionum in plano descriptione. Lugd. Batavor. 1646.

P. Gregorii a Sto. Vincentis, Opus Geometricum Quadratura Circuli et Sectionum Coni decem libris comprehensum. Antverpiae, 1647.

Johannis Wallisii De Sectionibus Conicis nova methodo expositis tractatus. Oxonii, 1655.

Joannis de Witt Elementa Curvarum Linearum per Franciscum Schooten edita cum Cartesii Geometria. Amstelodami, 1659 et 1661.

Euclides Adaeus et Methodicus Mathematicaque Universalis. Autore Guarino. Augustae Taurin. 1671.

Nouvelle méthode en Géométrie pour les Sections des superficies Coniques et Cylindriques qui ont pour base des cercles, ou des paraboles, des ellipses, et des hyperboles. Par Ph. de la Hire. Paris, 1673.

Nouveaux Éléments des Sections Coniques, &c. par M. de la Hire. Paris, 1679.

Sectiones Conicae in novem libr. distributae, &c. Aut. Phil. de la Hire. Paris, 1685.

Traité des Lignes du Premier Genre, expliquées par une Méthode nouvelle et facile. par M. Ozanam. Paris, 1687.

Sectionum Conicarum Elementa nova methodo demonstrata. Autore Jacobo Mihnes. Oxoniae, 1702.

Traité Analytique des Sections Coniques, et de leur usage, &c. Ouvrage posthume de M. le Marquis de l'Hôpital. Paris, 1707.

Compendio delle Sezioni Coniche d'Apollonio di P. Guido Grandi. Firen. 1722.

Delle Sezioni Coniche dedotte nuovamente in Piano del cerchio di Vincent. Santini. Luca, 1722.

Sectionum Conicarum Elementa methodo facillime demonstrata. Autore L. Trevigar. Cantabrigiae, 1731.

A Treatise on Conic Sections, by R. Steel. Dublin, 1723.

Elementa Sectionum Conicarum. Autore Nicolas de Martino. Tomii ii. Neapoli, 1734.

Elements of Conic Sections, in three books, by R. Jack. Edinburgh, 1742.

A Mathematical Treatise, containing a System of Conic Sections, &c. by J. Muller. London, 1736.

Sectionum Conicarum Elementa. Autore Josepho Bosovich. This treatise forms part of his Elementa Universale Matheseos. Romae, 1754.

De Sectionibus Conicis tractatus geometricus, in quo ex natura ipsius Coni Sectionum affectiones facillime deducuntur methodo nova. Autore Hugone Hamilton. Dublin, 1758. There is an English translation of this excellent work.

Sectionum Conicarum libri quinque. Autore R. Simson. Edinburgi, 1735 and 1750.

Antonii Rochii Conicarum Sectionum nova methodo expositarum specimen de proportionum compositione syntagma. Patavii, 1756.

Introduction aux Sections Coniques, par M. Mauduit. Paris, 1761.

Sectionum Conicarum Compendium. Autore D. Octaviano Cametti. Venetiis, 1765.

The Elements of the Conic Sections, in three books, by W. Emerson. London, 1767.

The Elements of the Conic Sections, as preparatory to the reading of Newton's Principia, by the Rev. S. Vince. Cambridge, 1781.

Elements of Conic Sections, &c. by Charles Hutton. LL.D. 1787.

Sectionum Conicarum Libri Septem. Accedit Tractatus de Sectionibus Conicis, et de Scriptoribus qui eorum doctrinam tradiderunt. Autore Abramo Robertson, A. M. Oxonii, 1792.

A Geometrical Treatise on Conic Sections, in four books, &c. by the Rev. Abram Robertson. Oxford, 1802.

A Short Treatise on the Conic Sections, in which the three Curves are derived from a general description on a Plane, by the Rev. T. Newton. Cambridge, 1794.

A System of Conic Sections adapted to the Study of Natural Philosophy, by the Rev. D. M. Peacock. 1810.

A Compendious and Practical Treatise on the Construction, Properties, and Analogies of the Three Conic Sections, by the Rev. D. Bridge. 1811.

Essai de Géométrie Analytique appliqué aux Courbes et aux Surfaces du second ordre, par J. B. Biot. Paris, 1810.

A Treatise on Lines of the Second Order. This is part of a work entitled Geometry of Curve Lines, by Sir John Leslie, formerly Professor of Mathematics, afterwards of Natural Philosophy, in the University of Edinburgh. Edinburgh, 1821.

A Treatise on the Conic Sections, in five books, by G. Walker, F.R.S. London, 1794.

The Elements of the Ellipse, together with the Radii of Curvature, by James Adams. London, 1818.

An Analytical System of the Conic Sections, by the Rev. H. P. Hamilton. Cambridge, 1828.

A Geometrical Treatise on the Conic Sections, by William Wallace, Professor of Mathematics in the University of Edinburgh. Edinburgh, 1833.

There is much valuable matter relating to the conic sections in several works which do not treat expressly on the subject; particularly in Newton's Principia, lib. i. The learned Jesuits Le Seur and Jacquier have given a concise treatise in their commentary on the work, at Prop. 8, lib. i. Maclaurin has treated of the conic sections in his Geometria Organica, sect. 1; in his Fluxions, chap. xiv. and in sect. 2 of the appendix to his Algebra. Euler has treated of them in his Introductio in Analysin Infinitorum, lib. ii. cap. 5; and De Moivre in his Miscellanea Analytica, lib. viii. cap. 2. The Synopsis Palmariorum Matheseos of Jones also treats of the subject; but it would extend our catalogue too much to name all the writers who have improved the theory. The reader may see a copious list of them in Bibliotheca Mathematica, Auctore Fred. Guil. Aug. Murhard. Lipsiae, 1798.

I. If a straight line BC, and a point without it F, be given in position in a plane, and a point D be supposed to move in such a manner that DF, its distance from the given point, is equal to DB, its distance from the given line; the point D will describe a line DAD', called a Parabola.

II. The straight line BC, which is given in position, is called the Directrix of the parabola.

III. The given point F is called the Focus.

IV. A straight line perpendicular to the directrix, terminated at one extremity by the parabola, and produced indefinitely within it, is called a Diameter.

V. The point in which a diameter meets the parabola is called its Vertex.

VI. The diameter which passes through the focus is called the Axis of the parabola; and the vertex of the axis is called the Principal Vertex.

COROLLARY. A perpendicular drawn from the focus to the directrix is bisected at the vertex of the axis.

VII. A straight line terminated both ways by the parabola, and bisected by a diameter, is called an Ordinate to that diameter.

VIII. The segment of a diameter between its vertex and an ordinate, is called an Absciss.

IX. A straight line quadruple the distance between the vertex of a diameter and the directrix, is called the Parameter, also the Latus Rectum of that diameter.

X. A straight line meeting the parabola only in one point, and which everywhere else falls without it, is said to touch the parabola at that point, and is called a Tangent to the parabola.

The distance of any point without the parabola from the focus is greater than its distance from the directrix; and the distance of any point within the parabola from the focus is less than its distance from the directrix.

Let DAD' be a parabola, of which F is the focus, GC the directrix, and P a point without the curve, that is, on the same side of the curve with the directrix; PF, a line drawn to the focus, will be greater than PG, a perpendicular to the directrix. For, as PF must necessarily cut the curve, let D be the point of intersection; draw DB perpendicular to the directrix, and join PB. Because D

is a point in the parabola, DB = DF (Definition 1), therefore PF = PD + DB; but PD + DB is greater than PB (20, 1, E), and therefore still greater than PG (19, 1, E), therefore PF is greater than PG.

Again, let Q be a point within the parabola; QF, a line drawn to the focus, is less than QB, a perpendicular to the directrix. The perpendicular QB necessarily cuts the curve; let D be the point of intersection; join DF. Then DF = DB (Def. 1), and QD + DF = QB; but QF is less than QD + DF, therefore QF is less than QB.

Cor. A point is without or within the parabola, according as its distance from the focus is greater or less than its distance from the directrix.

Every straight line perpendicular to the directrix meets the parabola, and every diameter falls wholly within it. (See Figure to Prop. I.)

Let the straight line BQ be perpendicular to the directrix at B; BQ shall meet the parabola. Draw BF to the focus, and make the angle BFP equal to FBQ; then, because QBC is a right angle, QBF and PFB are each less than a right angle, therefore QB and PF intersect each other; let D be the point of intersection, then DB = DF (5, 1, E); therefore D is a point in the parabola. Again, the diameter DQ falls wholly within the parabola; for take Q any point in the diameter, and draw FQ to the focus, then QB or QD + DF is greater than QF; therefore Q is within the parabola (Cor. 1).

Cor. The parabola continually recedes from the axis, and a point may be found in the curve that shall be at a greater distance from the axis than any assigned line.

The straight line which bisects the angle contained by two straight lines drawn from any point in the parabola, the one to the focus, and the other perpendicular to the directrix, is a tangent to the curve in that point.

Let D be any point in the curve; let DF be drawn to the focus, and DB perpendicular to the directrix; the straight line which bisects the angle FDB is a tangent to the curve. Join BF meeting DE in I, take H any other point in DE, join HF, HB, and draw HG perpendicular to the directrix. Because DF = DB, and DI is common to the triangles DFI, DBI, and the angles FDI, BDI, are equal, these triangles are equal, and FI = IB; and hence FH = HB (4, 1, E); but HB is greater than HG (19, 1, E); therefore the distance of the point H from the focus is greater than its distance from the directrix, hence that point is without the parabola (Cor. 1), and therefore HDI is a tangent to the curve at D (Def. 10).

Cor. 1. A perpendicular to the axis at its vertex is a tangent to the curve. Let AM be perpendicular to the axis at the vertex A, then RS, the distance of any point in AM from the directrix, is equal to CA, that is to AF,

Parabola, and therefore is less than $RF$, the distance of the same point from the focus.

Cor. 2. A straight line drawn from the focus of a parabola perpendicular to a tangent, and produced to meet the directrix, is bisected by the tangent. For it has been shown that $FB$, which is perpendicular to the tangent $DI$, is bisected at $I$.

Cor. 3. A tangent to the parabola makes equal angles with the diameter which passes through the point of contact, and a straight line drawn from that point to the focus. For $BD$ being produced to $Q$, $DQ$ is a diameter, and the angle $HDQ$ is equal to $BDE$, that is, to $EDF$.

Cor. 4. The axis is the only diameter which is perpendicular to a tangent at its vertex. For the angle $HDQ$, or $BDE$, is the half of $BDF$, and therefore less than a right angle, except when $BD$ and $DF$ lie in a straight line, which happens when $D$ falls at the vertex.

Cor. 5. There cannot be more than one tangent to the parabola at the same point.

For let any other line $DK$, except a diameter, be drawn through $D$; draw $FK$ perpendicular to $DK$; on $D$ for a centre, with a radius equal to $DB$ or $DF$, describe a circle, cutting $FK$ in $N$; draw $NL$ parallel to the axis, meeting $DK$ in $L$, and join $FL$. Then $FK\; =\; KN$ (3, 3, E.), and therefore $FL\; =\; LN$. Now $BD$ being perpendicular to the directrix, the circle $FBN$ touches the directrix at $B$ (16, 3, E.); and hence $N$, any other point in the circumference, is without the directrix, and on the same side of it as the parabola; therefore the point $L$ is nearer to the focus than to the directrix, and consequently is within the parabola.

SCHOLIUM. From the property of tangents to the parabola demonstrated in Cor. 3, the point $F$ takes the name of the Focus. For rays of light proceeding parallel to the axis of a parabola, and falling upon a polished surface whose figure is that produced by the revolution of the parabola about its axis, are reflected to the focus.

PROP. IV. PROBLEM.

To find any number of points in a parabola, having given the focus and axis.

Let $F$ be the focus, $AH$ the axis, and $A$ the vertex. Suppose the problem resolved, and that $D$ is a point in the parabola. In $FA$ produced take $AC$ equal to $AF$, and

through $C$ draw the directrix $BC$: draw $DF$ to the focus, $DE$ perpendicular to the axis, and $DB$ perpendicular to the directrix: Take $AH$ equal to $FD$.

Because $AH$ is equal to $DF$, and $DF$ is greater than $AF$ (Cor. Prop. 2), therefore $AH$ is greater than $AF$, and $H$ is always in $AF$ produced.

Now $CE$ is equal to $AH$, for each is equal to $DF$; therefore, taking from these the equals $AC$, $AF$, we have $AE\; =\; FH$.

CONSTRUCTION. In $AF$ produced take any point $H$, and take $AE$ equal to $FH$. Through $E$ draw $DE$ perpendicular to the axis, and with $F$ as a centre, at the distance $AH$, describe a circle which will cut the perpendicular in $D$ and $d$: these are points in the parabola. For $AE\; =\; FH$, therefore $CE\; =\; AH$, and $DB\; =\; DF$, therefore $D$ is in the parabola, and in the same way it appears that $d$ is in the parabola.

Cor. 1. Any perpendicular to the axis meets the parabola in two points, and in no more, and the straight line between the points is bisected by the axis. For if the perpendicular could meet the curve in another point $D\text{'}$, then $FD\text{'}$ being joined, we would have $FD\text{'}$ equal to $DB$, that is, to $DB$ or to $FD$, which is impossible (19, 1, E.).

Cor. 2. Every chord $Dd$, in a parabola, perpendicular to the axis, is bisected by the axis, and therefore is an ordinate to it. For the chord in the parabola is also a chord in a circle, the centre of which is in the axis of the parabola.

SCHOLIUM. From this proposition it appears that the parabola is composed of two branches, which recede continually from the directrix and from the focus, also from the axis (Cor. Prop. 2). And it appears that the indefinite spaces between the curve and axis on each side are exactly alike, so that if the whole space comprehended within the parabola were divided into two portions by cutting it through the axis, and one of them were turned over upon the other, they would entirely coincide.

PROP. V.

If a straight line be drawn from the focus of a parabola to the intersection of two tangents to the curve; it makes equal angles with straight lines drawn from the focus to the points of contact.

Let $HP$, $Hp$, tangents to a parabola at $P$ and $p$, intersect each other at $H$; draw $PF$, $pF$, $HF$, to $F$ the focus; the line $HF$ makes equal angles with $FP$, $Fp$.

Draw $PK$, $pK$ perpendicular to the directrix, and join $HK$, $Hk$. The triangles $HPK$, $HPF$ have $PK\; =\; PF$, $PH$ common to both, and the angles $KPH$, $FPH$ equal (3), therefore they are in every way equal (4, 1, E.), and have $HK\; =\; HF$, and the angle $HKP$ equal to $HFP$. In the same way it may be shown, that the triangles $Hpk$, $HpF$, are in every way equal, and therefore $Hk\; =\; HF$, and the

angle $HKp$ is equal to $HFp$: But $HK$ being equal to $Hk$, for each has been proved equal to $HF$, the angles $HKk$, $HAK$ are equal (5, 1, E.), and adding to these the right angles $PKk$, $pkk$, the angles $HKP$, $Hkp$ are equal; but these have been proved equal to $HFP$, $HFp$; therefore these last are equal, and the line $HF$ makes equal angles with $FP$, $Fp$.

COR. 1. Perpendiculars drawn from the intersection of two tangents, to lines drawn from the points of contact through the focus, are equal. For $HI$, $Hi$ being drawn perpendicular to $PF$, $Pf$, the triangles $HFI$, $HFi$ are manifestly equal (26, 1, E.), and therefore $HI\; =\; Hi$.

COR. 2. Perpendiculars from the intersection of two tangents to diameters passing through the points of contact are equal.

Draw $GHg$ through $H$ perpendicular to $PK$, $pk$, and because the triangles $HPG$, $HPI$ have $HP$ common to both, the angles at $P$ equal, and the angles $G$ and $I$ right angles, the triangles are in every way equal (26, 1, E.), and hence $HG$ is equal to $HI$. In like manner it is proved that $Hg$ is equal to $Hi$; but $HI$ is equal to $Hi$, therefore $HG$ is equal to $Hg$.

If a straight line be drawn from the intersection of two tangents to the focus, and another perpendicular to the directrix; these will make equal angles with the tangents.

Let $F$ be the focus of a parabola, and $Kk$ the directrix; and let straight lines $HP$, $Hp$, which intersect each other at $H$, touch the parabola at $P$ and $p$, also let $HF$ be drawn to the focus, and $HE$ perpendicular to the directrix; the angles $PHE$, $pHF$ are equal.

The same construction being made as in Prop. V. In the triangles $HEK$, $HEk$, it may be shown, as in that proposition, that $HK$ is equal to $Hk$; and therefore that the angle $HKE$ is equal to the angle $HkE$ (5, 1, E.). The angles $HEK$, $HEk$ are also equal; therefore the angles $KHE$, $kHE$ are equal (26, 1, E.). Now the angle $KHE\; =\; KHP\; +\; PHE$; but the triangles $KHP$, $FHP$ are in every way equal (as was shown in Prop. V.) Therefore $KHP\; =\; FHP$, and hence

In the same way it appears that

therefore $FHE\; +\; 2PHE\; =\; FHE\; +\; 2FHp$; and hence $2PHE\; =\; 2FHp$ and $PHE\; =\; FHp$.

If two tangents to a parabola be at the extremities of a chord, and a third tangent be parallel to the chord; the

part of this tangent intercepted between the other two Parabola, is bisected at the point of contact.

Let $HD$, $hd$ be tangents at the extremities of the chord $Dd$, and $KPk$ a tangent parallel to $Dd$, meeting the other tangents in $K$ and $k$; the line $Kk$ is bisected at $P$, the point of contact.

From $H$, $K$, $k$, the intersections of the tangents, draw perpendiculars to the diameters passing through their points of contact, viz. $HI$, $Hi$, perpendicular to $DL$ and $dl$; and $KM$, $KN$, perpendicular to $DL$ and $PE$, and $km$, $kn$, perpendicular to $dl$ and $PE$.

The triangles $DHI$, $DKM$, are manifestly equiangular, also the triangles $dHi$, $dkn$; therefore

But because $Kk$ is parallel to $Dd$,

therefore $HI\; :\; KM\; =\; Hi\; :\; kn$.

Now $HI\; =\; Hi$ (2 Cor. 5), therefore $KM\; =\; kn$.

But $KM\; =\; KN$, and $Kn\; =\; kn$ (2 Cor. 5);

therefore $KN\; =\; kn$.

And since $KN\; :\; kn\; =\; KP\; :\; kP$ (Cor. 6, E.),

therefore $KP\; =\; kP$.

Let $KLl$ be a triangle, having its base $Ll$ bisected at $p$, and let $Ha$, any straight line parallel to the base, and terminated by the sides, be bisected at $P$; then $P$, $p$, the points of bisection, and $K$, the vertex of the triangle, are in the same straight line; and that line bisects $Dd$, any other straight line parallel to the base.

Complete the parallelograms $KHPM$, $KLpN$. The triangles $KHh$, $KLl$ being similar, and $Ha$, $Ll$ similarly divided at $P$ and $p$,

hence the parallelograms $KHPM$, $KLpN$ are similar. Now they have a common angle at $K$, therefore they are about the same diameter, that is, the points $K$, $P$, $p$ are in the same straight line (26, 6, E.).

Next, let $Dd$ meet $Kp$ in $E$, then

therefore $DE$ is equal to $Ed$.

Any chord parallel to a tangent is bisected by the diameter which passes through the point of contact, or is an ordinate to that diameter.

The chord $Dd$, which is parallel to the tangent $KPk$, is bisected at $E$ by $PE$, the diameter that passes through the point of contact.

Let $DH$, $dH$ be tangents, and $DN$, $dn$ diameters at the extremities of the chord; let the tangent at $P$ meet the other tangents in $K$ and $k$, and the diameters in $L$ and $l$, and through $H$ draw $OHo$ parallel to $Dd$, and $IHi$ perpendicular to the diameters $DN$, $dn$.

Because of the parallels $Ll$, $Oo$, and $DO$, $do$, the triangles $DKL$, $DHO$ are similar, also the triangles $dkl$, $dHo$, and the triangles $OHI$, $oHi$, therefore

But because $Dd$ is parallel to $Kk$,

therefore $KL\; :\; HO\; =\; kl\; :\; Ho$;

therefore, ex aq. $KL\; :\; HI\; =\; kl\; :\; Hi$.

Now $HI\; =\; Hi$ (2 Cor. 5), therefore $KL\; =\; kl$; but $KP\; =\; kp$ (7), therefore $PL\; =\; Pl$, and $ED\; =\; Ed$ (34, 1, E.)

COR. 1. Straight lines which touch a parabola at the extremities of an ordinate to a diameter intersect each other in that diameter; for $Kk$ and $Dd$ being bisected at $P$ and $E$, the points $H$, $P$, $E$ lie in a straight line. (LEMMA.)

COR. 2. Every ordinate to a diameter is parallel to a tangent at its vertex: For if it be not, let a tangent be drawn parallel to the ordinate; then the diameter which passes through the point of contact would bisect the ordinate, and thus the same line would be bisected in two different points, which is impossible.

COR. 3. All ordinates to the same diameter are parallel to each other.

COR. 4. A straight line that bisects two parallel chords, and terminates in the curve, is a diameter.

COR. 5. The axis is perpendicular to its ordinates, and every other diameter cuts its ordinates obliquely.

If a tangent at any point in a parabola meet a diameter, and from the point of contact an ordinate be drawn to that diameter; the segment of the diameter between the vertex and tangent is equal to the segment between the vertex and the ordinate.

Let $DH$, a tangent to the curve at $D$, meet the diameter $EP$ in $H$, and let $DEd$ be an ordinate to that diameter: the segment $HE$ is bisected in $P$.

Draw $PK$, a tangent at $P$, meeting the tangent $DH$ in $K$, and draw $IKi$, perpendicular to the diameter $PE$ at $i$, and meeting a diameter drawn through $D$ at $l$.

The triangles $DKI$, $HKi$ are similar (29, 1, E.), therefore $IK\; :\; Ki\; =\; DK\; :\; KH$ (4, 6, E.); and because in the triangle $DHE$, $KP$ is parallel to the side $DE$, $DK\; :\; KH\; =\; EP\; :\; PH$, therefore $IK\; :\; Ki\; =\; EP\; :\; PH$; but $IK$ and $Ki$ are equal (2 Cor. 5), therefore $EP$ and $PH$ are equal.

If an ordinate to any diameter pass through the focus; the absciss is equal to one fourth of the parameter of that diameter, and the ordinate is equal to the whole parameter.

Let $DEd$, a straight line passing through the focus, be an ordinate to the diameter $PE$; the absciss $PE$ is equal to one fourth the parameter, and the ordinate $Dd$ is equal to the whole parameter of the diameter $PE$.

Let $DH$, $PI$ be tangents at $D$ and $P$; let $DH$ meet the diameter in $H$; draw $PF$ to the focus, and $DL$ parallel to $EP$. The angles $HPI$, $IPF$, being equal (3), and $PI$ parallel to $EF$ (2 Cor. 8), the angles $PEF$, $PFE$, are also equal (29, 1, E.), and $PE\; =\; PF\; =\; \backslash frac\{1\}\{2\}$ the parameter (Def. 9 and Def. 1). Again, the angle $HDE$ is equal to $LDH$ (3), and therefore equal to $DHE$; consequently $ED$ is equal to $EH$, or to twice $EP$ (9): therefore $Dd$ is equal to $4EP$, or to $4PF$, that is, to the parameter of the diameter.

If any two diameters of a parabola be produced to meet a tangent to the curve; the segments of the diameters between their vertices and the tangent are to one another as the squares of the segments of the tangent intercepted between each diameter and the point of contact.

Let $QH$, $RK$, any two diameters, be produced to meet $PI$, a tangent to the curve at $P$, in the points $G$, $I$; then, $HG\; :\; KI\; =\; PG^2\; :\; PI^2$.

Let $PN$, a semi-ordinate to the diameter $HQ$, meet $KR$ in $O$, and let $PR$, a semi-ordinate to the diameter $KO$, meet $HN$ in $Q$; from $H$ draw parallels to $NO$ and $QR$, meeting $KR$ in $L$ and $M$; thus $HL$ is a tangent to the curve, and $HM$ a semi-ordinate to $KR$.

Now $KI\; =\; KR$, and $KL\; =\; KM$ (9); therefore, by subtraction, $LI\; =\; MR\; =\; HQ$.

But $LO\; =\; HN\; =\; HG$ (9); therefore, by addition, $IO\; =\; GQ$.

The triangles $PGN$, $PIO$, are similar, as also $PGQ$, $PIR$,

therefore $GN$, or $2GH\; :\; IO\; =\; PG\; :\; PI$,

and $GQ\; :\; IR$, or $IO\; :\; 2IK\; =\; PG\; :\; PI$.

Hence, taking the rectangles of the corresponding terms,

$2GH\; \backslash cdot\; IO\; :\; 2IO\; \backslash cdot\; IK\; =\; PG^2\; :\; PI^2$,

therefore $GH\; :\; IK\; =\; PG^2\; :\; PI^2$.

Cor. The squares of semi-ordinates, and of ordinates to any diameter, are to one another as their corresponding abscisses.

Let $HEA$, $KNK$ be ordinates to the diameter $PN$; draw $PG$ a tangent to the curve at the vertex of the diameter, and complete the parallelograms $PEHG$, $PNKI$; then $PG$, $PI$, are equal to $EH$, $NK$, and $GH$, $IK$, to $PE$, $PN$, respectively; therefore $HE^2\; :\; KN^2\; =\; PE\; :\; PN$.

If an ordinate be drawn to any diameter of a parabola; the rectangle under the absciss and the parameter of the diameter is equal to the square of the semi-ordinate.

Let $KBA$ be an ordinate to the diameter $PB$; the rectangle contained by $PB$, and the parameter of the diameter is equal to the square of $KB$, the semi-ordinate.

Let $DEd$ be that ordinate to the diameter which passes through the focus. The semi-ordinates $DE$, $Ed$ are each half of the parameter, and the absciss $EP$ is one fourth of the parameter (10);

therefore $Dd\; :\; DE\; =\; DE\; :\; PE$,

and $Dd\; \backslash cdot\; PE\; =\; DE^2$ (16, 6, E.).

But $Dd\; \backslash cdot\; PE\; :\; Dd\; \backslash cdot\; PB$, or $PE\; :\; PB\; =\; DE^2\; :\; KB^2$ (Cor. 9);

therefore $Dd\; \backslash cdot\; PB\; =\; KB^2$.

SCHOLIUM. It was on account of the equality of the square of the semi-ordinate to a rectangle contained by the parameter of the diameter and the absciss, that Apol-

lonius called the curve line to which the property belonged a Parabola.

If $AB$, an ordinate to a diameter $PG$, is cut by any other diameter $CD$ in $D$; the rectangle $AD\; \backslash cdot\; DB$ contained by its segments is equal to the rectangle contained by $CD$, the segment of the diameter between its vertex and the ordinate, and the parameter of the diameter $PG$.

Draw $CH$, a semi-ordinate to the diameter $PG$, and let $L$ be its parameter.

Because $AG^2\; =\; L\; \backslash cdot\; PG$ (12),

and $DG^2\; =\; CH^2\; =\; L\; \backslash cdot\; PH$,

therefore $AG^2\; -\; DG^2\; =\; L(PG\; -\; PH)$,

that is (5 and 6, 2, E.), $AD\; \backslash cdot\; DB\; =\; L\; \backslash cdot\; CD$.

Cor. If a chord $AB$ is cut by any two diameters $CD$, $EF$, the rectangles $AD\; \backslash cdot\; DB$, $AF\; \backslash cdot\; FB$, are to one another as $CD$, $EF$, the segments of the diameters between their vertices and the chord.

For since $AD\; \backslash cdot\; DB\; =\; L\; \backslash cdot\; CD$; and $AF\; \backslash cdot\; FB\; =\; L\; \backslash cdot\; EF$; therefore $AD\; \backslash cdot\; DB\; :\; AF\; \backslash cdot\; FB\; =\; L\; \backslash cdot\; CD\; :\; L\; \backslash cdot\; EF\; =\; CD\; :\; EF$.

A straight line drawn from the focus of a parabola, perpendicular to a tangent, is a mean proportional between the straight line drawn from the focus to the point of contact, and one fourth the parameter of the axis.

Let $FB$ be a perpendicular from the focus upon the tangent $PB$, and $FP$ a straight line drawn to the point of contact; let $A$ be the principal vertex, and therefore $FA$ equal to one fourth of the parameter of the axis; $FB$ is a mean proportional between $FP$ and $FA$.

Produce $FB$ and $FA$ to meet the directrix in $D$ and $C$, and join $AB$. The lines $FC$, $FD$ are bisected at $A$ and $B$ (2 Cor. 3), therefore (2, 6, E.) $AB$ is parallel to $CD$, or perpendicular to $CF$, and consequently is a tangent to the curve at $A$ (1 Cor. 3). Now $BP$ is a tangent at $P$, therefore the angle $AFB$ is equal to $BFP$ (5); and since the angles $FAB$, $FBP$ are right angles, the triangles $FAB$, $FBP$ are equiangular; hence

$FP\; :\; FB\; =\; FB\; :\; FA$.

Cor. 1. The common intersection of a tangent, and a perpendicular from the focus to the tangent, is in a straight line touching the parabola at its vertex.

Cor. 2. If PH be drawn perpendicular to the tangent, meeting the axis in H, and HK be drawn perpendicular to PF; PK shall be equal to half the parameter of the axis. For the triangles HPK, FBP, are manifestly equiangular; therefore

But if PD be joined, the line PD is evidently perpendicular to the directrix (8), therefore the figure HPDF is a parallelogram, and $HP\; =\; FD$, therefore $PK\; =\; FC\; =$ half the parameter of the axis.

If from a point in a parabola a perpendicular be drawn to any diameter, and also, from the same point, an ordinate to that diameter: the square of the perpendicular is equal to the rectangle contained by the absciss of the diameter and the parameter of the axis.

Let P be a point in a parabola, DK any diameter, PH a perpendicular, and PK a semi-ordinate to that diameter; the square of PH is equal to the rectangle contained by DK and the parameter of the axis.

Let F be the focus, and FA the segment of the axis between the focus and vertex, and therefore one fourth of the parameter; join FD, draw DB touching the parabola at D, and FB a perpendicular from the focus on the tangent. The triangles PKH, FDB are similar, for the angle FDB is equal to BDH (3 Cor. 3), that is, to PKH (2 Cor. 8), and the angles at B and H are right angles, therefore their sides are proportional,

But since $DF\; :\; FB\; =\; FB\; :\; FA$ (14),

therefore $KP^2\; :\; PH^2\; =\; DF\; :\; FA\; =\; 4\; DF\; :\; DK\; :\; 4\; FA\; :\; DK$. Now $KP^2\; =\; 4\; DF\; :\; DK$ (12), therefore $PH^2\; =\; 4\; FA\; :\; DK$.

Cor. 1. Hence, if two diameters DK, dk, on opposite sides of a third PQ, be at equal distances from it, semi-ordinates PK, Pk to the other two, drawn from P the vertex of the middle diameter, will cut off equal abscisses DK, dk. For the perpendiculars PH, Ph, on the two extreme diameters, are equal.

Cor. 2. And if tangents DI, dI, be drawn at the vertices of the extreme diameters, they will intersect each other in the middle diameter PQ produced. For the tangents being parallel to the ordinates (8), each will cut off from PQ a segment PI equal to the absciss of the diameter at the point of contact; and the abscisses DK, dk being equal, the tangents will cut off equal segments from PQ, and therefore will pass through the same point I.

Cor. 3. And if two diameters be at equal distances from a third, on opposite sides, and chords be drawn from the vertex of the middle diameter to the vertices of the other two, tangents drawn parallel to the chords will in-

tersect each other in the middle diameter produced. For the semi-ordinates PK, Pk, are the halves of chords so drawn, and DI, dI are tangents parallel to these chords.

Let CD, any diameter of a parabola whose vertex is C, intersect a chord AB in D; from the ends of the chord inflect straight lines AE, BE, to E, any point in the curve, and let these cut the diameter in H and K, the point H being in AE, and K in BE; the segments AD, BD of the chord shall have the same ratio as the segments CH, CK of the diameter between its vertex and the inflected lines.

From A, either extremity of the chord, draw AF parallel to the diameter CD, meeting BE in F.

By similar triangles, $BF\; :\; BK\; =\; BA\; :\; BD$, and $FE\; :\; KE\; =\; AF\; :\; HK$;

therefore, taking the rectangles of corresponding terms of the ratios,

But (13) $BF\; \backslash cdot\; FE\; :\; BK\; \backslash cdot\; KE\; =\; AF\; :\; KC\; =\; BA\; \backslash cdot\; AF\; :\; BA\; \backslash cdot\; KC$ (1, 6, E);

Cor. Let BA be any chord in a parabola, and BL a tangent to the curve at one extremity of the chord; let any straight line DCL parallel to the axis meet the chord in D, the curve in C, and the tangent in L; the chord AB and the line DL will be similarly divided at D and C, that is, $AD\; :\; DB\; =\; DC\; :\; CL$.

Draw chords to E, any point in the curve, and let them meet DL in H and K: by the proposition $AD\; :\; DB\; =\; HC\; :\; CK$.

Suppose now that the point E moves along the curve, until at last it comes to B, the point of contact of the tangent; the line BK will then become BL, and AH will become AD, and the ratio of CH to CK will become the ratio of CD to CL; therefore $AD\; :\; DB\; =\; CD\; :\; CL$.

The directrix and focus of a parabola being given by position, to describe the parabola by a mechanical construction.

Let AB be the given directrix, and F the focus. Place the edge of the ruler ABKH along the directrix, and keep it fixed in that position. Let LCG be another ruler of such a form that the side LC may slide along AB, the edge of the fixed ruler ABKH, and the part CG may have its edge CD constantly perpendicular to AB. Let GDF be a string of the same length as GC, the edge of the moveable ruler; let one end of the string be fixed at F, and the other fastened to G, a point in the moveable ruler. By means of the pin D let the string be stretched so that the part of it between G and D may be applied close to the edge of the moveable ruler, while at the same time the ruler slides along BA, the edge of the fixed ruler; the pin D will thus be constrained to move along CG, the edge of the ruler, and its point will trace upon the plane in which the directrix and focus are situated, a curve line DE, which is the parabola required. For the string GDF being equal in length to GDC, if GD be taken from both, there remains DF equal to DC; that is, the distance of the moving point D from the focus is equal to its distance from the directrix, therefore the point D describes a parabola.

A parabola being given by position, to find its directrix and focus.

Let $Dd$ be the given parabola; draw any two parallel chords $Dd$, $Ee$, and bisect them at $H$ and $K$; join $KH$, meeting the parabola in $P$; the straight line $PHK$ is a diameter (4 Cor. 8), the point $P$ is its vertex, and $Dd$, $Ee$ are ordinates to it. In $HP$ produced take $PL$ equal to one fourth part of a third proportional to $PH$ and $HD$, and draw $LN$ perpendicular to $PL$, the line $LN$ will evidently be the directrix (12, and Def. 9). Draw $PM$ parallel to the ordinates to the diameter $PK$, then $PM$ will be a tangent to the curve at $P$ (2 Cor. 8). Draw $LM$ perpendicular to $PM$, and take $MF\; =\; ML$, and the point $F$ will be the focus of the parabola (3 Cor. 3).

I. If two points $F$ and $f$ be given in a plane, and a point $D$ be conceived to move around them in such a manner that $Df\; +\; DF$, the sum of its distances from them, is always the same; the point $D$ will describe upon the plane a line $Abab$, which is called an ellipse.

II. The given points $F$, $f$ are called the foci of the ellipse.

III. The point $C$ which bisects the straight line between the foci is called the centre.

IV. The distance of either focus from the centre is called the eccentricity.

V. A straight line passing through the centre, and terminated both ways by the ellipse, is called a diameter.

VI. The extremities of a diameter are called its vertices.

VII. The diameter which passes through the foci is called the transverse axis, also the greater axis.

VIII. The diameter which is perpendicular to the transverse axis is called the conjugate axis, also the lesser axis.

IX. Any straight line not passing through the centre, but terminated both ways by the ellipse, and bisected by a diameter, is called an ordinate to that diameter.

X. Each of the segments of a diameter intercepted between its vertices and an ordinate, is called an absciss.

XI. A straight line which meets the ellipse in one point only, and everywhere else falls without it, is said to touch the ellipse in that point, and is called a tangent to the ellipse.

If from any point in an ellipse two straight lines be drawn to the foci, their sum is equal to the transverse axis.

Let $Abab$ be an ellipse, of which $F$, $f$ are the foci, and $Aa$ the transverse axis; let $D$ be any point in the curve, and $DF$, $Df$ lines drawn to the foci; $Df\; +\; DF\; =\; Aa$.

Because $A$, $a$ are the points in the ellipse,
$Af\; +\; AF\; =\; af\; +\; af$ (Def. 1),
therefore $Af\; +\; 2AF\; =\; af\; +\; 2af$;
hence $2AF\; =\; 2af$, and $AF\; =\; af$,
and $Af\; +\; AF\; =\; Af\; +\; af\; =\; Aa$.
But $D$ and $A$ being points in the ellipse,
$Df\; +\; DF\; =\; Af\; +\; AF$,
therefore $Df\; +\; DF\; =\; Aa$.

Cor. 1. The sum of two straight lines drawn from a point without the ellipse to the foci is greater than the transverse axis. And the sum of two straight lines drawn from a point within the ellipse to the foci is less than the transverse axis.

Let $PF$, $Pf$ be drawn from a point without the ellipse to the foci; let $Pf$ meet the ellipse in $D$; join $FD$; then $Pf\; +\; PF$ is greater than $Df\; +\; DF$ (21, 1, E.), that is, than $Aa$. Again, let $QF$, $Qf$ be drawn from a point within the ellipse, let $Qf$ meet the curve in $D$, and join $FD$; $Qf\; +\; QF$ is less than $Df\; +\; DF$ (21, 1, E.), that is, than $Aa$.

Cor. 2. A point is without or within the ellipse, according as the sum of two lines drawn from it to the foci is greater or less than the transverse axis.

Cor. 3. The transverse axis is bisected in the centre. Let C be the centre, then $CF\; =\; Cf$ (Def. 3), and $FA\; =\; fa$, therefore $CA\; =\; Ca$.

Cor. 4. The distance of either extremity of the conjugate axis from either of the foci is equal to half the transverse axis. Let $Bb$ be the conjugate axis; join $Fb,\; fb$. Because $CF\; =\; Cf$, and $Cb$ is common to the triangles $CFb,\; Cfb$, also the angles at C are right angles, these triangles are equal; hence $Fb\; =\; fb$, and since $Fb\; +\; fb\; =\; Aa$, $Fb\; =\; AC$.

Cor. 5. The conjugate axis is bisected in the centre. Join $fb,\; fB$. By the last corollary $Bf\; =\; bf$; therefore the angles $fBC,\; fbC$ are equal; now $fC$ is common to the triangles $fCB,\; fCb$, and the angles at C are right angles, therefore (26, 1, E.) $CB\; =\; Cb$.

Every diameter of an ellipse is bisected in the centre.

Let $Pp$ be a diameter, it is bisected in C. For if $Cp$ be not equal to $CP$, take $CQ$, equal to $CP$, and from the points $P,\; p,\; Q$, draw lines to $F,\; f$, the foci. The triangles $FCP,\; fCQ$ having $FC\; =\; Cf,\; PC\; =\; CQ$, and the angles at C equal, are in all respects equal, therefore $FP\; =\; fQ$; in like manner it appears that $fP\; =\; FQ$, therefore $FQ\; +\; fQ$ is equal to $FP\; +\; fP$, or (Def. 1) to $Fp\; +\; fp$, which is absurd (21, 1, E.); therefore $CP\; =\; Cp$.

Cor. 1. Every diameter meets the ellipse in two points only.

Cor. 2. Every diameter divides the ellipse into two parts, which are equal and similar, the like parts of the curve being at opposite extremities of the diameter.

The square of half the conjugate axis of an ellipse is equal to the rectangle contained by the segments into which the transverse axis is divided by either focus.

Draw a straight line from $f$, either of the foci, to B, either of the extremities of the conjugate axis.

Then $BC^2\; +\; Cf^2\; =\; Bf^2\; =\; Ca^2$ (4 Cor. 1).

But because $Aa$ is bisected at C,

To find any number of points in an ellipse, having given the transverse axis and foci.

Let $F,\; f$ be the foci, $Aa$ the transverse axis, and C the centre. Suppose the problem resolved, and that D is a point in the ellipse; join $DF,\; Df$; take $AH$ in the axis equal to $DF$; then $aH$ will be equal to $Df$ (1).

And $HA\; -\; Ha\; =\; DF\; -\; Df$;

But $HA\; -\; Ha\; =\; HC\; +\; Ca\; -\; Ha\; =\; 2CH$; therefore $DF\; -\; Df\; =\; 2CH$.

Now ;

and hence ;
therefore and .

Thus it appears that the point H may be anywhere between the foci; but that it cannot be between the foci and the vertices.

CONSTRUCTION.—Take H, any point between the foci, and from F and f as centres, with the distances $HA,\; Ha$ describe circles, which will cut each other in two points D, d, one on each side of the axis. These are points in the ellipse.

Join $DF,\; Df$, also $dF,\; df$. Because $DF\; +\; Df\; =\; HA\; +\; Ha\; =\; Aa$, therefore D is a point in the ellipse. In like manner, it appears that d is in the ellipse.

In this way, by taking different points H, may any number of points be found.

Cor. 1. Any perpendicular to the transverse axis between its extremities meets the ellipse in two points, and in no more. For, if the perpendicular $Dd$ could meet the curve in two points D, D', on the same side of the axis, then $DF,\; Df$, also $D\text{'}F,\; D\text{'}f$, being drawn to the foci, $DF\; +\; Df$ would be equal to $D\text{'}F\; +\; D\text{'}f$. Now, supposing D to be the point nearer to the axis, $DF$ will be greater than $D\text{'}F$, and $Df$ greater than $D\text{'}f$ (19, 1, E.), and $DF\; +\; Df$ greater than $D\text{'}F\; +\; D\text{'}f$; therefore D and D' cannot both be points in the ellipse.

Cor. 2. Every chord $Dd$, in an ellipse, perpendicular to the transverse axis, is bisected by that axis, and therefore is an ordinate to it. For the chord $Dd$ in the ellipse is also a chord in a circle, the centre of which is in the axis.

Cor. 3. Of all the straight lines that can be drawn from either focus to the curve, the longest is that which passes through the centre; and the shortest is the remainder of the transverse axis. And only two equal straight lines can be drawn from the focus to the curve, one on each side of the axis.

SCHOLIUM. From this proposition it appears that an ellipse is a curve, which returns into itself, thus inclosing a finite area; also, that the spaces between the curve and the axis on each side are alike in every way; so that if the ellipse were resolved into two portions, by cutting it along the axis, the space $ABDa$, if turned over, would coincide entirely with the space $Abda$. Now it has been shown that the same spaces will coincide if one of them be reversed (2 Cor. 2); and then the curve $ABDa$ will coincide with $adbA$. Hence it follows that the two axes divide the whole ellipse into four portions exactly alike; and which, by superposition, may be applied on each other.

The straight line which bisects the angle adjacent to that which is contained by two straight lines drawn from any point in the ellipse to the foci, is a tangent to the curve in that point.

Let D be any point in the curve; let DF, Df be straight lines drawn to the foci; the straight line DE which bisects the angle fDG adjacent to fDF, is a tangent to the curve at D.

Take H any other point in DE; take DG = Df, and join Hf, HF, HG, fG; let fG meet DE in L. Because Df = DG, and DL is common to the triangles DfL, DGL, and the angles fDL, GDL are equal, these triangles are equal, and fL = LG, and hence fH = HG (4, 1, E.), and FH + fH = FH + HG; but FH + HG is greater than FG, that is, greater than FD + fD or Aa; therefore FH + fH is greater than Aa; hence the point H is without the ellipse (2 Cor. 1); and therefore DHI is a tangent to the curve at D (Def. 11).

COR. 1. There cannot be more than one tangent at the same point; for D is such a point in the line DE, that the sum of DF, Df, the distances of that point from the foci, is evidently less than the sum of HF, Hf, the distances of H, any other point in that line; and if another line KDI be drawn through D, there is in like manner a point K in that line which will be different from D, such, that the sum of FK, fK is less than the sum of the distances of any other point in KI, and therefore less than FD + fD; therefore the point K will be within the ellipse (2 Cor. 1), and the line KI will cut the curve.

COR. 2. A perpendicular to the transverse axis at either of its extremities is a tangent to the curve. The demonstration is the same as for the proposition, if it be considered that when D falls at either extremity of the axis, the point L falls also at the extremity of the axis; and thus the tangent DE, which is always perpendicular to fL, is perpendicular to the axis.

COR. 3. A perpendicular to the conjugate axis at either of its extremities is a tangent to the curve. For the perpendicular evidently bisects the angle adjacent to that which is contained by lines drawn from the extremity to the foci.

COR. 4. A tangent to the ellipse makes equal angles with straight lines drawn from the point of contact to the foci. For the angle fDE being equal to GDE, is also equal to FDM, which is vertical to GDE.

SCHOLIUM. From the property of the ellipse which forms this last corollary, the points F and f take the name of Foci. For writers on optics show that if a polished concave surface be formed, whose figure is that produced by the revolution of an ellipse about its transverse axis, rays of light which flow from one focus, and fall upon that surface, are reflected to the other focus; so that if a luminous point be placed in one focus, there is formed by reflection an image of it in the other focus.

The tangents at the vertices of any diameter of an ellipse are parallel.

Let Pp be a diameter, and HPK, hpk tangents at its vertices; draw straight lines from P and p to F and f the foci. The triangles FCP, fCp, having FC = fC, CP = Cp (2), and the angles at C equal, are in all respects equal; and because the angle FPC is equal to Cpf, FP is parallel to fp (27, 1, E.); therefore Pf is equal and parallel to pF (33, 1, E.), thus FPfp is a parallelogram, of which the opposite angles P and p are equal (34, 1, E.). Now the angles FPH, fph are evidently half the supplements of these angles (4 Cor. 5), therefore the angles FPH, fph, are equal, and hence CPH, Cph are also equal, and consequently HP is parallel to hp.

COR. 1. If tangents be drawn to an ellipse at the vertices of a diameter; straight lines drawn from either focus to the points of contact make equal angles with these tangents. For the angle Fpk is equal to FPH.

COR. 2. The axes of an ellipse are the only diameters which are perpendicular to tangents at their vertices. For let Pp be any other diameter, then PF and pf are necessarily unequal, and therefore the angles FpP, FpP are also unequal; to these add the equal angles Fpk, FPH, and the angles Cpk, CPH are unequal; therefore neither of them can be a right angle (20, 1, E.).

If a straight line be drawn from either focus of an ellipse to the intersection of two tangents to the curve; it will make equal angles with straight lines drawn from the same focus to the points of contact.

Let HP, Hp, tangents to an ellipse at P and p, intersect each other at H; draw PF, pf, HF to F, either of the foci; the line HF makes equal angles with FP, Fp.

Draw Pf, pf to f the other focus, and in FP, Fp produced take PK = Pf and pk = pf; join HK, Hk, and draw fK, fh, meeting the tangents in G and g.

The triangles HPK, HPf have PK = Pf, PH common to both, and the angles KPH, fPH equal (4), therefore they are in every way equal (4, 1, E.), and have HK = Hf. In the same way it may be shown that the triangles Hpk, Hpf are in every way equal, and therefore that Hk = Hf.

The triangles HFK, HFk have HK = Hk (for each is equal to Hf), HF common to both, and FK = Fk, because each is equal to PF + Pf or pF + pf, that is, to the trans-

verse axis; therefore they are in all respects equal, and the angle $HFK$ is equal to the angle $HFk$; wherefore $HF$ makes equal angles with $FP$ and $Fp$.

Cor. Perpendiculars drawn from the intersection of two tangents to straight lines drawn from either focus through the points of contact are equal.

Let $HI,\; Hi$ be perpendiculars drawn from $H$, the intersection of the tangents $PH,\; p\backslash hat\{h\}$ on the lines $FP,\; Fp$. The triangles $HFI,\; Hfi$ are in all respects equal (26, 1, E.), therefore $HI\; =\; Hi$.

Straight lines drawn from the intersection of two tangents to the foci, make equal angles with the tangents.

Let $F,\; f$ be the foci of an ellipse, and let straight lines $HP,\; Hp$, which intersect each other at $H$, touch the ellipse at $P$ and $p$, also let $HF,\; Hf$ be lines drawn to the foci; the angles $PHF,\; p\backslash hat\{H\}f$ are equal.

The same construction being made as in Prop. VI., because the angles $FHK,\; FHk$ are equal,

and in like manner,

If two tangents to an ellipse be at the extremities of a chord, and a third tangent be parallel to the chord; the part of this tangent intercepted by the other two is bisected at the point of contact.

Let $HD,\; Hd$ be tangents at the extremities of the chord $Dd$, and $Kpk$ a tangent parallel to $Dd$, meeting the other tangents in $K$ and $k$; the intercepted segment $Kk$ is bisected at $P$, the point of contact.

From the points of contact $D,\; P,\; d$, draw lines to $F$,

either of the foci, and from $H,\; K,\; k$, the intersections of the tangents, draw perpendiculars to the lines drawn from the points of contact to the foci, viz. $HI,\; Hi$ perpendicular to $DF,\; dF$; and $KM,\; KN$ perpendicular to $FD,\; FP$; and $km,\; kn$ perpendicular to $Fd,\; FP$.

The triangles $DHI,\; DKM$ are manifestly equiangular, also the triangles $dHi,\; dkm$;

but because $Dd$ is parallel to $Kk$, a side of the triangle $HKA$,

therefore $HI\; :\; KM\; =\; Hi\; :\; km$.

Now $HI\; =\; Hi$ (Cor. 7), therefore $KM\; =\; km$; but $KM\; =\; KN$ and $km\; =\; kn$ (1 Cor. 7), therefore $KN\; =\; kn$; and since from the similar triangles $KPN,\; kPn$, $KN\; :\; kn\; =\; KP,\; kP$, therefore $KP$ is equal to $kP$.

Any chord not passing through the centre, but parallel to a tangent, is bisected by the diameter that passes through the point of contact; or it is an ordinate to that diameter.

The chord $DEd$, which is parallel to $Kk$, a tangent at $P$, is bisected at $E$ by the diameter $PCp$.

Draw $Lpl$, a tangent at $p$, the other end of the diameter, and $DH,\; dH$, tangents at $D$ and $d$, the extremities of the chord, meeting the other tangents in $K,\; k$, and $L,\; l$. Then $KPk$ and $Lpl$ are bisected at $P$ and $p$ (9); therefore the diameter $Pp$, when produced, will pass through $H$, and bisect $Dd$, which is parallel to $Kk$ or $Ll$, in $E$. (Lemma to Prop. 8, Part I.)

Cor. 1. Straight lines which touch an ellipse in the extremities of an ordinate to any diameter, intersect each other in that diameter.

Cor. 2. Every ordinate to a diameter is parallel to a tangent at its vertex: for if not, let a tangent be drawn parallel to the ordinate; then the diameter drawn through the point of contact would bisect the ordinate; and thus the same line would be bisected in two different points, which is absurd.

Cor. 3. All the ordinates to the same diameter are parallel to each other.

Cor. 4. A straight line that bisects two parallel chords, and terminates in the curve, is a diameter.

Cor. 5. The ordinates to either axis are perpendicular to that axis; and no other diameter is perpendicular to its ordinates.

If a tangent to an ellipse meet a diameter, and from the point of contact an ordinate be drawn to that diameter; the semidiameter will be a mean proportional between the segments of the diameter intercepted between the centre and the ordinate, and between the centre and the tangent.

Let $DH$, a tangent to the ellipse at $D$, meet the diameter $Pp$ produced in $H$, and let $DEd$ be an ordinate to that diameter: $CE\; :\; CP\; =\; CP\; :\; CH$.

Through $P$ and $p$, the vertices of the diameter, draw the tangents $PK$, $pL$, meeting $DH$ in $K$ and $L$; draw $PF$, $pF$, to either of the foci, and draw $KM$ and $KN$ perpendicular to $FD$ and $FP$, also $LO$ and $LI$ perpendicular to $FD$ and $Fp$.

The triangles $PKN$, $pLI$, are equiangular; for the angles at $N$ and $I$ are right angles, and the angles $NPK$, $IpL$ are equal (1 Cor. 6); therefore $PK\; :\; pL\; =\; KN\; :\; LI$ (4, 6, E.) $=\; KM\; :\; LO$ (Cor. 7). But the triangles $KDM$, $LDp$, being manifestly equiangular,

therefore $PK\; :\; pL\; =\; KD\; :\; LD$.

But because of the parallel lines $PK$, $ED$, $pL$, the triangles $HPK$, $HpL$, are equiangular; and the lines $HL$, $Hp$, are similarly divided in $K$, $D$, and in $P$, $E$ (10, 6, E.), hence

therefore $HP\; :\; Hp\; =\; PE\; :\; pE$.

Take $CG\; =\; CE$, and then $PE\; =\; pG$, and by conversion

and taking the halves of the consequents,

and by composition, $HC\; :\; PC\; =\; PC\; :\; EC$.

Cor. 1. The rectangle $PE\; \backslash cdot\; Ep$ is equal to the rectangle $HE\; \backslash cdot\; EC$.

For $PC^2\; =\; HC\; \backslash cdot\; CE$ (17, 6, E.)

therefore $HE\; \backslash cdot\; EC\; =\; PE\; \backslash cdot\; Ep$.

Cor. 2. The rectangle $PH\; \backslash cdot\; Hp$ is equal to the rectangle $EH\; \backslash cdot\; HC$.

For $HC^2\; =\; PH\; \backslash cdot\; Hp\; +\; CP^2$ (6, 2, E.);

and $HC^2\; =\; EH\; \backslash cdot\; HC\; +\; EC\; \backslash cdot\; HC$ (1, 2, E.);

therefore $PH\; \backslash cdot\; Hp\; =\; EH\; \backslash cdot\; HC$.

If a diameter of an ellipse be parallel to the ordinates to another diameter; the latter diameter is parallel to the ordinates to the former.

Let $Qq$, a diameter of an ellipse, be parallel to $DEd$, any ordinate to the diameter $Pp$; the diameter $Pp$ shall be parallel to the ordinates to the diameter $Qq$. Ellipse.

Draw the diameter $dCd$ through one extremity of the ordinate $Dd$, and join $d$ and $D$, the other extremity of the ordinate, meeting $Qq$ in $G$. Because $dD$ is bisected at $G$ (2, 6, E.); therefore $Dd$ is an ordinate to the diameter $Qq$ (Def. 9), and because $dD$ and $Dd$ are bisected at $C$ and $E$, the diameter $Pp$ is parallel to $Dd$ (2, 6, E.); therefore $Pp$ is parallel to any ordinate to the diameter $Qq$.

XII. Two diameters are said to be conjugate to one another when each is parallel to the ordinates to the other diameter.

Cor. Diameters which are conjugate to one another are parallel to tangents at the vertices of each other.

XIII. A third proportional to any diameter and its conjugate is called the Parameter, also the Latus Rectum of that diameter.

If an ordinate be drawn to any diameter of an ellipse; the rectangle contained by the segments of the diameter will be to the square of the semi-ordinate as the square of the diameter to the square of its conjugate.

(See Figure to Prop. XII.)

Let $DEd$ be an ordinate to the diameter $Pp$, and let $Qq$ be its conjugate.

Let $KDL$, a tangent at $D$, meet the diameter in $K$, and its conjugate in $L$; draw $DG$ parallel to $Pp$, meeting $Qq$ in $G$. Because $CP$ is a mean proportional between $CE$ and $CK$ (11),

and, by division, $CP^2\; :\; PE\; \backslash cdot\; Ep\; =\; CK\; :\; KE$.

But because $ED$ is parallel to $CL$,

and because $CQ$ is a mean proportional between $CG$ and $CL$ (11),

therefore $CP^2\; :\; PE\; \backslash cdot\; Ep\; =\; CQ^2\; :\; DE^2$,

and, by inversion and alternation,

Cor. 1. The squares of semi-ordinates and of ordinates to any diameter of an ellipse are to one another as the rectangles contained by the corresponding abscisses.

Cor. 2. The ordinates to any diameter, which intercept equal segments of that diameter from the centre, are equal to one another; and, conversely, equal ordinates intercept equal segments of the diameter from the centre.

Cor. 3. If a circle be described upon $Aa$, either of the axes of an ellipse, as a diameter, and $DE$, $de$, any two semi-ordinates to the axes, meet the circle in $H$ and $h$, $DE$ shall be to $de$ as $HE$ to $he$.

For $DE^2\; :\; dc^2\; =\; AE\; \backslash cdot\; Ea\; :\; Ae\; \backslash cdot\; ea\; =\; HE^2\; :\; he^2$,
therefore $DE\; :\; dc\; =\; HE\; :\; he$.

COR. 4. If a circle be described on $Aa$ the transverse axis as a diameter, and $DE$, any ordinate to the axis, be produced to meet the circle in $H$; $HE$ shall be to $DE$ as the transverse axis $Aa$ to the conjugate axis $Bb$. For, produce the conjugate axis to meet the circle in $K$, then, by last corollary,

$HE\; :\; DE\; =\; KC$, or $AC\; :\; BC\; =\; Aa\; :\; Bb$.

COR. 5. And if $HE$ be divided at $D$, so that $HE$ is to $DE$ as the transverse axis to the conjugate axis, $D$ is a point in the ellipse, and $DE$ a semi-ordinate to the axis $Aa$.

In $Bb$, the conjugate axis of an ellipse, let there be taken on each side of the centre $C$, straight lines $CK$, $Ck$, each a fourth proportional to $CF$, the eccentricity, and $CA$, $CB$, half the transverse and conjugate axes: If then from $P$, a vertex of any diameter, there be drawn $PH$ perpendicular to $Bb$; the square of the semi-diameter $PC$ will have to the rectangle contained by the segments $KH$, $kH$, the constant ratio of the square of $CF$ to the square of $CB$.

Draw $PL$ perpendicular to the transverse axis.

Because (13 of this, and 5, 2, E.)

and, by division,

therefore ($\backslash frac\{1\}\{4\}$ Cor. 1, and 47, 1, E.)

But, by hypothesis, $CF^2\; :\; CB^2\; =\; CA^2\; :\; CK^2$;

therefore $CA^2\; :\; CK^2\; =\; CA^2\; -\; PC^2\; :\; CH^2$,

and hence (19, 5, E.)

and (5, 2, E.)

COR. 1. Hence the squares of any semidiameters $PC$, $QC$, are to one another as the rectangles $KH\; \backslash cdot\; Hk$, $KI\; \backslash cdot\; Ik$, contained by the segments of the line $Kk$, between its extremities, and perpendiculars from the vertices of the diameters.

COR. 2. The transverse axis is the greatest diameter, and the conjugate axis the least; and a diameter which is nearer to the transverse axis is greater than one more remote.

By hypothesis $CF\; :\; CA\; =\; CB\; :\; CK$, therefore $CK$ is greater than $CB$, and the points $K$, $k$, are without the ellipse: Suppose now a semidiameter $PC$ to turn about $C$, and that in every position $PH$ is perpendicular to $Kk$: The rectangle $KH\; \backslash cdot\; Hk$ will manifestly be greatest when $PC$ coincides with $AC$, and least when it coincides with $BC$, and will decrease continually while $PC$ passes from the position $AC$ to $BC$; therefore the same will be true of the revolving semidiameter $PC$, which has a constant ratio to the rectangle $KH\; \backslash cdot\; Hk$.

COR. 3. Diameters which make equal angles with the transverse axis on opposite sides of it are equal; and only two equal diameters can be drawn, one on each side of the transverse axis.

If an ordinate be drawn to any diameter of an ellipse, the rectangle under the abscisses of the diameter is to the square of the semi-ordinate as the diameter to its parameter.

Let $DE$ be a semi-ordinate to the diameter $Pp$, let $PG$ be the parameter of the diameter, and $Qq$ the conjugate diameter. By the definition of the parameter (Def. 13),

therefore $Pp\; :\; PG\; =\; Pp^2\; :\; Qq^2$ (2 Cor. 20, 6, E.).

therefore $PE\; \backslash cdot\; Ep\; :\; DE^2\; =\; Pp\; :\; PG$.

COR. Let the parameter $PG$ be perpendicular to the diameter $Pp$; join $pG$, and from $E$ draw $EM$ parallel to $PG$, meeting $pG$ in $M$. The square of $DE$, the semi-ordinate, is equal to the rectangle contained by $PE$ and $EM$.

therefore $DE^2\; =\; PE\; \backslash cdot\; EM$.

SCHOLIUM. If the rectangles $PGI\_p$, $HGKM$, be completed, it will appear that the square of $ED$ is equal to the rectangle $MP$, which rectangle is less than the rectangle $KP$, contained by the absciss $PE$ and parameter $PG$, by a rectangle $KH$ similar and similarly situated to $LP$, the rectangle contained by the diameter and parameter. It was on account of the deficiency of the square of the ordinate from the rectangle contained by the absciss and parameter that Apollonius called the curve line to which the property belonged an ellipse.

If from the vertices of two conjugate diameters of an ellipse there be drawn ordinates to any third diameter; the square of the segment of that diameter intercepted between either ordinate and the centre is equal to the rectangle contained by the segments between the other ordinate and the vertices of the same diameter.

Let $Pp,\; Qq$ be two conjugate diameters, and $PE,\; QG$ semi-ordinates to any third diameter $Rr$; $CG^2\; =\; RE\; \backslash cdot\; Er$, and $CE^2\; =\; RG\; \backslash cdot\; Gr$.

Draw the tangents $PH,\; QK$ meeting $Rr$ in $H$ and $K$. The rectangles $HC\; \backslash cdot\; CE$ and $KC\; \backslash cdot\; CG$ are equal, for each is equal to $CR^2$ (11), therefore

But the triangles $HPC,\; CQK$ are evidently similar (Cor. Def. 12), and $PE$ being parallel to $QG$, their bases $CH,\; KC$ are similarly divided at $E$ and $G$, therefore

wherefore $CG\; :\; CE\; =\; HE\; :\; EG$, consequently $CG^2\; =\; CE\; \backslash cdot\; EH\; =\; (1\; \backslash text\{\; Cor.\; 10\})\; RE\; \backslash cdot\; Er$. In like manner it may be shown that $CE^2\; =\; RG\; \backslash cdot\; Gr$.

COR. 1. Let $Ss$ be the diameter that is conjugate to $Rr$, then $Rr$ is to $Ss$ as $CG$ to $PE$, or as $CE$ to $QG$.

In like manner $Rr\; :\; Ss\; ::\; CE\; :\; QG$.

COR. 2. The sum of the squares of $CE,\; CG$, the segments of the diameter to which the semi-ordinates $PE,\; QG$ are drawn, is equal to the square of $CR$ the semi-diameter.

COR. 3. The sum of the squares of any two conjugate diameters is equal to the sum of the squares of the axes.

Let $Rr,\; Ss$ be the axes, and $Pp,\; Qq$ any two conjugate diameters; draw $PE,\; QG$ perpendicular to $Rr$, and $PL,\; QM$ perpendicular to $Ss$. Then

If four straight lines be drawn touching an ellipse at the vertices of any two conjugate diameters; the parallelogram formed by these lines is equal to the rectangle contained by the transverse and conjugate axes.

Let $Pp,\; Qq$ be any two conjugate diameters; a parallelogram $DEGH$ formed by tangents to the curve at their vertices is equal to the rectangle contained by $Aa,\; Bb$, the two axes.

Produce $Aa$, one of the axes, to meet the tangent $PE$ in $K$; join $QK$, and draw $PL,\; QM$ perpendicular to $Aa$.

But $CK\; \backslash cdot\; QM\; =\; \backslash text\{twice\; trian.\; \}\; CKQ\; =\; \backslash text\{paral.\; \}\; CPEQ\; \backslash text\{\; (41,\; 1,\; E.),\}$

and taking the quadruples of these, the parallelogram $DEGH$ is equal to the rectangle contained by $Aa$ and $Bb$.

If two tangents at the vertices of any diameter of an ellipse meet a third tangent; the rectangle contained by their segments between the points of contact and the points of intersection is equal to the square of the semidiameter to which they are parallel. And the rectangle contained by the segments of the third tangent between its point of contact and the parallel tangents is equal to the square of the semidiameter to which it is parallel.

Let $PH,\; ph$, tangents at the vertices of a diameter $Pp$, meet $HDh$, a tangent to the curve at any point $D$, in $H$ and $h$: let $CQ$ be the semidiameter to which the tangents $PH,\; ph$ are parallel, and $CR$ that to which $HDh$ is parallel;

If the tangent $HDh$ be parallel to $Pp$, the proposition is manifest. If it be not parallel, let it meet the semi-diameters $CP,\; CQ$, in $L$ and $K$. Draw $DE,\; RM$ parallel to $CQ$, and $DG$ parallel to $CP$.

hence, and because of the parallels $PH,\; ED,\; CK,\; ph$,

Again, the triangles $LED,\; CMR$ are evidently similar, and $LE,\; LD$ similarly divided at $H$ and $P$, also at $h$ and $p$, therefore $PE\; :\; HD\; =\; (LE\; :\; LD)\; =\; CM\; :\; CR$,

hence, taking the rectangles of the corresponding terms,

But if $CD$ be joined, the points $D$ and $R$ are evidently the vertices of two conjugate diameters (Cor. Def. 12), and therefore $PE\; \backslash cdot\; pE\; =\; CM^2$ (16):

COR. The rectangle contained by $LD$ and $DK$, the segments of a tangent intercepted between $D$, the point of contact, and $Pp,\; Qq$, any two conjugate diameters, is equal to the square of $CR$, the semidiameter to which the tangent is parallel.

Let the parallel tangents $PH,\; ph$ meet $LK$ in $H$ and $h$, and draw $DE$ a semi-ordinate to $Pp$. Because of the parallels $PH,\; ED,\; CK,\; ph$,

If two straight lines be drawn from the foci of an ellipse perpendicular to a tangent; straight lines drawn from

the centre to the points in which they meet the tangent will each be equal to half the transverse axis.

Let $Dd$ be a tangent to the curve at $P$, and $FD,\; fd$ perpendiculars to the tangent from the foci; the straight lines joining the points $C,\; D$, and $C,\; d$, are each equal to $AC$, half the transverse axis.

Join $FP,\; fP$, and produce $FD,\; fP$ till they intersect in $E$. The triangles $FDP,\; EDP$ have the angles at $D$ right angles, and the angles $FDP,\; EDP$ equal (4), and the side $DP$ common to both; they are therefore equal, and consequently have $ED\; =\; DF$, and $EP\; =\; PF$, therefore $Ef\; =\; FP\; +\; Pf\; =\; Aa$. Now the straight lines $FE,\; Ef$, being bisected at $D$ and $C$, the line $DC$ is parallel to $Ef$, and thus the triangles $FfE,\; FCD$ are similar;

therefore $Ff\; :\; fE$ or $Aa\; =\; FC\; :\; CD$.

But $FC$ is half of $Ef$, therefore $CD$ is half of $Aa$. In like manner it may be shown that $Cd$ is half of $Aa$.

COR. If the diameter $Qq$ be drawn parallel to the tangent $Dd$, it will cut off from $PF,\; Pf$ the segments $PG,\; Pg$, each equal to $AC$ half the transverse axis. For $CdPG,\; CDPg$, are parallelograms, therefore $PG\; =\; dC\; =\; AC$, and $Pg\; =\; DC\; =\; AC$.

The rectangle contained by perpendiculars drawn from the foci of an ellipse to a tangent is equal to the square of half the conjugate axis.

(See Figure to Prop. XIX.)

Let $Dd$ be a tangent, and $FD,\; fd$ perpendiculars from the foci; the rectangle contained by $FD$ and $fd$ is equal to the square of $CB$ half the conjugate axis.

It is evident from the last proposition that the points $D,\; d$ are in the circumference of a circle whose centre is the centre of the ellipse, and radius $CA$ half the transverse axis; now $FDd$ being a right angle, if $dC$ be joined, the lines $DF,\; dC$, when produced, will meet at $H$, a point in the circumference; and since $FC\; =\; fC$, and $CH\; =\; Cd$, and the angles $FCH,\; fCd$ are equal, $FH$ is equal to $fd$; therefore

COR. If $PF,\; Pf$ be drawn from the point of contact to the foci, the square of $FD$ is a fourth proportional to $fP,\; FP$ and $BC^2$. For the lines $fP,\; FP$ make equal angles with the tangent (4 Cor. 4), and $fdP,\; FDP$ are right angles, therefore the triangles $fPd,\; FPD$ are similar, and

If from $C$ the centre of an ellipse a straight line $CL$ be drawn perpendicular to a tangent $LD$, and from $D$ the point of contact a perpendicular be drawn to the tangent, meeting the transverse axis in $H$ and the conjugate axis in $h$; the rectangle contained by $CL$ and $DH$ is equal to the square of $CB$, the semi-conjugate axis; and the rectangle contained by $CL$ and $Dh$ is equal to the square of $CA$, the semi-transverse axis.

Produce the axes to meet the tangent in $M$ and $m$, and from $D$ draw the semi-ordinates $DE,\; De$, which will be perpendicular to the axes.

Produce the axes to meet the tangent in $M$ and $m$, and from $D$ draw the semi-ordinates $DE,\; De$, which will be perpendicular to the axes.

The triangles $DEH,\; CLm$ are evidently equiangular, therefore

In the same way it is shown that $CL\; \backslash cdot\; Dh\; =\; AC^2$.

COR. 1. If a perpendicular be drawn to a tangent at the point of contact, the segments intercepted between the point of contact and the axes are to each other reciprocally as the squares of the axes by which they are terminated.

COR. 2. If $DF$ be drawn to either focus, and $HK$ be drawn perpendicular to $DF$, the straight line $DK$ shall be equal to half the parameter of the transverse axis.

Draw $CG$ parallel to the tangent at $D$, meeting $DH$ in $N$, and $DF$ in $G$. The triangles $GDN,\; HDK$ are similar, therefore

But $GD\; =\; AC$ (Cor. 19), and $ND\; =\; CL$, therefore $AC\; \backslash cdot\; DK\; =\; HD\; \backslash cdot\; CL\; =$ (by the Prop.) $CB^2$;

wherefore $AC\; :\; BC\; =\; BC\; :\; DK$;

hence $DK$ is half the parameter of $Aa$ (Def. 13).

XIV. If a point $G$ be taken in the transverse axis of an ellipse produced, so that the distance of $G$ from the centre may be a third proportional to $CF$ the eccentricity, and $CA$ the semi-transverse axis; a straight line $HGh$, drawn through $G$ perpendicular to the axis, is called the directrix of the ellipse.

COR. 1. If $Mm$, an ordinate to the axis, be drawn through the focus; tangents to the ellipse at the extremities of the ordinate will meet the axis at the point $G$ (11).

COR. 2. The ellipse has two directrices, for the point G may be taken on either side of the centre.

The distance of any point in an ellipse from either directrix is to its distance from the focus nearest that directrix in the constant ratio of the semitransverse axis to the eccentricity.

Let D be any point in the ellipse, let DK be drawn perpendicular to the directrix, and let DF be drawn to the focus nearest the directrix; DK is to DF as CA, half the transverse axis, to CF, the eccentricity.

Draw Df to the other focus, and DE perpendicular to Aa; take L a point in the axis, so that AL = FD, and consequently La = Df; then CL is evidently half the difference between AL and aL, or FD and fD, and CE half the difference between fE and FE; and because $Df\; +\; DF\; :\; fF\; =\; fE\; -\; FE\; :\; Df\; -\; DF$ (K, and 16 of 6, E.),

By taking the halves of the terms of the proportion,

COR. 1. If the tangent GMN be drawn through M, the extremity of the ordinate passing through the focus, and ED be produced to meet GM in N, EN shall be equal to DF. For draw MO perpendicular to the directrix, then, because M and D are points in the ellipse,

But the triangles GFM, GEN being similar,

COR. 2. If AI and ai be drawn perpendicular to the transverse axis at its extremities, meeting the tangent GM in I and i, then AI = AF and ai = aF.

For GA : AF = OM : MF = GF : MF = GA : AI, therefore AF = AI; and, in like manner, it may be shown that aF = ai.

Two unequal straight lines which bisect each other at right angles being given by position; to describe an ellipse of which these may be the two axes, by a mechanical construction.

Let Aa be the transverse, and Bb the conjugate axes. About either extremity of the conjugate axis as a centre, with a radius equal to CA, half the transverse axis, describe arcs cutting that axis in F and f; these points will be the foci (4 Cor. 1). Let the ends of a string equal in length to Aa be fastened at the points F, f, and let the string be stretched by a pin at D, and while it is kept uniformly tense, let the point of the pin be carried along the plane about the centre C, till it return to the place from whence it set out. By this motion the point of the pin will trace on the plane a curve which will be the ellipse required, as is evident from the definition of the ellipse.

An ellipse being given by position; to find its axes.

Let ABab be the given ellipse: draw two parallel chords Hh, Kk, and bisect them at L and M; join LM, and produce it to meet the ellipse in P and p, then Pp is a diameter (4 Cor. 10). Bisect Pp in C; the point C is the centre of the ellipse (2).

Take D any point in the ellipse, and on C as a centre, with the distance CD, describe a circle. If this circle be wholly without the curve, then CD must be half the transverse axis; but if it be wholly within the curve, then CD must be half the conjugate axis (14). If the circle neither be wholly without nor wholly within the ellipse, let the circle meet it again in d. Join Dd, and bisect Dd in E; join CE, and produce CE to meet the ellipse in A and a; then Aa will be one of the axes (5 Cor. 10); for it is perpendicular to the line Dd (3, 3, E.), which is an ordinate to Aa: the other axis Bb will be found by drawing a straight line through the centre perpendicular to Aa.

If a moveable circle roll along the concave circumference of a fixed circle in the same plane, and the radius of the former be half that of the latter; any given point in its circumference will describe a diameter of the fixed circle.

Let ABHO a moveable circle (which may be called the generating circle), whose centre is C, roll along Dhd a

Ellipse. fixed circle, of which O is the centre, both being in the same plane, and let the radius CH be half of the radius OH; any given point A in the circumference of the generating circle will always be in Dd, some diameter given in position of the other circle.

Let H be the point of contact of the circles: the points H, C, O are in a straight line (11, 3, E.); and because the diameter of the inner circle is half that of the outer circle, one of its extremities will always be at O, the centre of the fixed circle.

Suppose that at the beginning of the motion the point A was at D, a given point in the circumference of the fixed circle, and that by rolling along the arch DH, the generating circle has come to the position OAH: draw AC to its centre, and bisect the angle ACH, and consequently the arch AH, by the radius BC.

The arch DH is equal to the arch AH, because every element of the one has been applied to an equal corresponding element of the other: therefore the arch DH is double the arch BH, and the radius of the circle DH is by hypothesis double the radius of the circle BH. Now in different circles, equal angles at their centres stand on arches which have the same ratio as their circumferences or their radii, therefore an angle at O, the centre of the fixed circle, standing on the arch DH, will be equal to the angle BCH, that is, to the angle AOH (20, 3, E.); hence a straight line drawn through O and A will pass through the given point D; thus A will always be in the diameter Dd, which is given by position, and by the motion of the circle will describe that diameter.

COR. 1. A diameter Dd drawn through the moving point A, in any one position, will be its Locus in every position.

COR. 2. The generating circle will have made two complete revolutions about its centre C when its diameter has completed one revolution about the centre O.

COR. 3. When the generating circle has made a complete revolution about O, every point in its circumference will have described a diameter, passing twice through the centre, and have returned to its first position.

SCHOLIUM. The refined notion of generating lines by supposing a curve to roll along a straight line or curve, is due to the moderns.

Galileo appears to have been the first who introduced it into geometry, and in this way he indicated the eyeloid, the discussion of which by Mersenne, Descartes, Pascal, and others, was the beginning of that series of discoveries which has since gradually expanded into the modern geometry. The ancients, however, in some cases employed motion in the formation of geometrical figures. Euclid defined a sphere to be the solid figure described by the revolution of a semicircle about its diameter, which remains unmoved; and Archimedes defined his spiral by the uniform motion of a point along a straight line which at the same time turns with an uniform angular motion about one of its extremities. The preceding proposition, and the following, may be established by the ordinary method of geometrical reasoning; or instead of supposing one circle to roll on another, we might suppose the diameter of the generating circle to turn about the centre of the fixed circle with a uniform angular motion, while at the same time it turned uniformly about its own centre, so as to make two complete revolutions in the time its diameter makes one. It then might be easily shown, that the extremity of any radius of the revolving circle would describe a diameter of a fixed circle.

The property here demonstrated is elegant, and remarkable in having been applied in mechanics to the production of a reciprocating rectilinear motion by means of a rotatory motion.

If a moveable or generating circle roll along the concave circumference of a fixed circle in the same plane, and the radius of the former be half that of the latter, as in the preceding proposition; any given point in the plane of the generating circle, within or without it, will describe an ellipse, of which conjugate diameters will be given in position.

Let DHEDE' be the fixed circle, O its centre, and AHBO the generating circle, which rolls along the concave circumference DHE: any given point P (the generating point) in the plane of this circle, within or without it, will describe an ellipse given in position.

Take a given point A anywhere in the circumference of the generating circle, and draw a straight line through A, and the generating point P, meeting the circumference in B: Thus A and B will be given points in the circle, and AP, PB lines given in magnitude. Draw OD, OE, radii of the fixed circle, through the points A and B: these will be lines given in position. (Preceding Prop.)

Let OCH be the revolving diameter of the generating circle, and C its centre. When OH has made a complete revolution about the centre O, the point A will have been twice at O, and BP, one of the segments of AB, will have coincided entirely with OM, OM', equal segments of the diameter DD', on opposite sides of the centre; therefore, if in DD' there be taken OM and OM', each equal to the line BP, M and M' will be given points in which the locus of P cuts the line DD'.

For a like reason AP, the other segment of AB, will have coincided with segments of the diameter EE' in two opposite positions, viz. ON, ON': therefore, if ON and ON' be taken each equal AP, N and N' will be also given points in which the locus of P intersects the line EE'.

By the motion of the generating circle, the point B, either of the extremities of the revolving chord AB, will in the course of a revolution have come to E, a point in the circumference of the fixed circle. The angle OAB will then be in a semicircle (of the generating circle), and AB will have the position of a perpendicular to OD. Let this be the line EF, which will be given in position, and also in magnitude, because it is equal to the chord BA. Take FQ = AP, therefore QE = PB; then Q will be a fifth given point in the locus of P. Join QO, and take OQ' = OQ: the line QQ', which is bisected at O, will be given in position and magnitude. Draw BK perpendicular

lar to OD, and PI parallel to it, meeting BK in I. The triangles BPI, BAK are similar (4, 6, E.), therefore

Because the lines BK, EF are similarly divided at Q and I, the points O, I, Q, are in a straight line (Lemma to Prop. 8, Part I.), so that I is in the line QOQ'.

Again, because $EO\; :\; OB\; =\; QO\; :\; OI$, and that OB cannot exceed OE, therefore OI cannot be greater than OQ; and when AB is not perpendicular to DD', the point I will always be between Q and Q': in no case can it be beyond these limits.

The triangles OQE, OIB, are similar (4, 6, E.), therefore their sides are proportional, and

and, by division, $OQ^2\; :\; QI\; \backslash cdot\; IQ\text{'}\; =\; BP^2\; :\; PI^2$;

hence, by alternation, and observing that $BP\; =\; OM$,

therefore the point P is in an ellipse, of which QQ' and MM', lines given in magnitude and position, are conjugate diameters (converse of 18).

Cor. 1. The ellipse described by the generating point P depends entirely on the magnitude of the generating circle AOB, and the position of the point in its plane, that is, on OC the radius of the circle, and CP its distance from the centre; therefore, whatever be the position of the revolving chord AB, if it always pass through the same point P, the ellipse will be the same, and have the same position on the plane of the fixed circle.

Cor. 2. The position of DD', EE', the diameters of the fixed circle, which are the loci of the extremities of the revolving chord AB, and consequently the position of MM' NN', the diameters of the ellipse, depend entirely on the position of the chord in respect of the centre; they will be different for different chords, but for the same chord they will have a fixed position. This is evident from the last proposition.

Cor. 3. The semidiameters OM, ON of the ellipse in which (produced if necessary) the revolving chord terminates, are equal to the distances of the generating point P from the ends of the chord, viz. OM to PB, and ON to PA, and the angle which the diameters MM', NN' make at the centre is half the angle which the chord subtends at the centre of the generating circle; for in the course of a revolution of the generating circle about the centre Q, the segments PB, PA of the chord will have been applied upon the lines OM, ON, so as entirely to coincide with them: the rest is evident (20, 3, E.).

Cor. 4. When the revolving chord AB comes into the position EF, a perpendicular to MM', either of the diameters, which are the loci of its extremities, it then passes through Q, a vertex of the diameter QOQ', which is the conjugate of MM', and in this position it is a normal to the ellipse: for a tangent to the ellipse at Q is parallel to MM', and therefore perpendicular to EF.

SCHOLIUM. From this proposition it appears that if the ends of a straight line AB of a given length be carried along two straight lines DOD', EOE', given in position; any point P in AB (or in AB produced), at given distances from its extremities, will describe an ellipse, the centre of which will be at the intersection of the lines given in position. It is upon this principle that elliptic compasses and lathes for turning ovals are constructed. An instrument for describing ellipses will be described farther on.

PROP. XXVII.

Supposing the ellipse MPN, whose centre is O, to have been described according to the hypothesis of Prop. XXVI.; let AOB be the generating circle in any position on the plane of the fixed circle, and P the generating point, which is also a point in the ellipse; through

P draw any chord APB; join OA and OB; take OM and OM' in contrary directions, each equal to PB, and ON and ON' in contrary directions, each equal to PA; then MM', NN' will be two diameters of the ellipse.

Since A and B are points in the circumference of the generating circle, and O is the centre of the fixed circle, the lines AO, BO will have the same position in the ellipse for all positions of the chord AB, because they are the loci of the points A, B; therefore (3 Cor. 26) the semidiameters OM, ON will be respectively equal to PB and PA, the distances of P from the ends of the chord AB; hence if OM, ON be taken equal to PA and PB, the points MN will be the vertices of diameters of the ellipse.

Cor. And if other chords ab, &c. be drawn through P, and Oa, Ob, &c. be joined, and there be taken Om equal to bP, and On equal to Pa, &c. then m, n, &c. will be points in the ellipse; and in this way any number of points whatever may be found from a single position of the generating circle.

PROP. XXVIII.

Supposing an ellipse to be described according to the hypothesis of Prop. XXVI. if the generating point be within the generating circle (fig. 1), half the sum of its semiaxes is equal to the radius of the circle; and half their difference, to the distance of the generating point from its centre. But if the point be without the circle (fig. 2), then half the difference of the semiaxes is equal to its radius, and half their sum to the distance of the generating point from its centre.

Fig. 1.

Let AOB be the generating circle (fig. 1 and 2), and P the generating point, at any point in the curve. It appears from Prop. XXVI. Cor. 4, that a chord in the circle passing through P and A, the intersection of the circle, and MM', one of the axes, will also pass through B, the point in which it meets the other axis NN'; and in this

Hyperbola. case the chord will pass through C the centre (31, 3, E.), because the axes form right angles at the centre of the ellipse. Therefore when the point P is within the circle AOB (fig. 1), the radius OC or AC is half the sum of BP and AP, that is, of OM and ON, the semiaxes (27); and CP, the distance of the generating point from the centre, is half the difference of PA and PB, or of OM and ON.

When the generating point P is without the circle AOB (fig. 2), then the radius OC or AC is manifestly half the difference of AP and BP; and CP, the distance of the generating point from the centre, is half the sum of AP and BP, that is, of OM and ON, the semiaxes.

COR. 1. Hence it appears that the same ellipse may be described by two different generating circles (each rolling on its own fixed circle), viz. by one whose diameter is the sum of the semiaxes, and by another whose diameter is their difference; in the first way the generating point will be within the circle, and in the second without it.

COR. 2. Also it appears that the chord of the generating circle, intercepted between the diameters of the ellipse that pass through its extremities (and which passes through the generating point), is equal to the sum of the semidiameters when the generating point is within the circle, but to their difference when the point is without the circle.

SCHOLIUM. The curves which may be generated by a point in the plane of a moveable circle which rolls along the circumference of a fixed circle, are called cycloids, also epicycloids. They are of two kinds, one generated by a circle rolling on the convex circumference, and another by its rolling on the concave. Some writers confine the name epicycloids to the first class, and call the second hypocycloids. It appears from this proposition that an ellipse is an hypocycloid. This property of the curve has not been observed, as far as we know, by any writer on

the conic sections, although the extreme case, viz. that in Hyperbola, which the ellipse, by the shortening of its lesser axis, degenerates into a straight line of a given length (Prop. 25), is always mentioned by writers on cycloids.

The property in question has suggested an instrument for generating an ellipse elegantly, by continued motion. A and B are two wheels, the axes of which turn in holes C, O, near the ends of the connecting bar D. One of the wheels B must be just half the diameter of the other A, which may be of any size, and a band EF goes round them, outside; an arm CP is attached to the wheel B, and admits of being lengthened or shortened by sliding along its surface in a socket which may be anywhere on the wheel. Suppose now that the wheel A is fixed or kept from turning, and that the bar D is turned round the centre O, carrying at its other extremity the wheel B; the action of the band EF will then turn this wheel B round its centre C, and while the bar makes one revolution round the centre of the fixed wheel, the other wheel will make two revolutions about its centre.

The use of the sliding arm CP is to give extension to the surface of the wheel, so that P, any point in the arm, may be regarded as a point in the plane of a circle turning about a moveable centre C, while that centre revolves about a fixed centre O. From this description it is easy to see that C, the centre of the wheel B, may be regarded as the centre of a circle which rolls on the inside of a circle whose centre is O; also that any point P in the plane which is the extension of the surface of the wheel, is just a point in the plane of the rolling circle; and since the circle of which C is the centre makes two turns in going round that of which O is the centre, the radius of the one circle must be double that of the other; and hence it follows from the proposition that the path of the point P in space is an ellipse.

I. If two points F, f be given in a plane, and a point D be conceived to move in such a manner that $Df\; -\; DF$, the difference of its distances from them, is always the same; the point D will describe upon the plane a line DAD' called an Hyperbola. By assuming first one of the given points F, and then the other f, as that to which the moving

point is nearest, the difference of the lines DF and Df in both cases being the same, there will be two hyperbolas DAD', dad' described, opposite to each other, which are therefore called Opposite Hyperbolas.

COR. The lines DF, Df may become greater than any given line, therefore the hyperbolas extend to a greater distance from the given points F, f than any which can be assigned.

II. The given points F, f are called the Foci of the hyperbola.

III. The point C, which bisects the straight line between the foci, is called the Centre.

IV. The distance of either focus from the centre is called the Eccentricity.

V. A straight line passing through the centre, and terminated by the opposite hyperbolas, is called a Transverse Diameter. It is also sometimes called simply a Diameter.

VI. The extremities of a diameter are called its Vertices.

VII. The diameter which passes through the foci is called the Transverse Axis.

Cor. The vertices of the transverse axis lie between the foci. Let A be either of the vertices, then, because any side of a triangle is greater than the difference between the other two sides, $Ff$ is greater than $fD\; -\; DF$, which is equal to $fA\; -\; FA$ (Def. 1). Now this can only be true when A is between F and f.

VIII. A straight line $Bb$ passing through the centre, perpendicular to the transverse axis, and limited at B and b by a circle described on one extremity of that axis, with a radius equal to the distance of either focus from the centre, is called the Conjugate Axis. It is also called the Second Axis.

Cor. The conjugate axis is bisected in the centre. This appears from 3, 3, E.

IX. Any straight line terminated both ways by the hyperbola, and bisected by a transverse diameter produced, is called an Ordinate to that diameter.

X. Each of the segments of a transverse diameter produced, intercepted by its vertices and an ordinate, is called an Absciss.

XI. A straight line which meets the hyperbola in one point only, and which everywhere else falls without the opposite hyperbolas, is said to touch the hyperbola in that point, and is called a Tangent to the hyperbola.

If from any point in an hyperbola two straight lines be drawn to the foci; their difference is equal to the transverse axis.

Let DA, da be opposite hyperbolas, of which F, f are the foci and Aa the transverse axis; let D be any point in the curve, and DF, Df lines drawn to the foci;

Because A and a are points in the hyperbola,

But D and A being points in the hyperbola,

Cor. 1. The difference of two straight lines drawn from a point without the opposite hyperbolas to the foci is less than the transverse axis, and the difference of two straight lines drawn from a point within either of them to the foci is greater than the transverse axis.

Let Pf, PF be lines drawn from a point without the hyperbolas, that is, between the curve and its conjugate axis. The line PF must necessarily meet the curve; let D be the point of intersection; Pf is less than $PD\; +\; Df$ (20, 1, E.), therefore $Pf\; -\; PF$ is less than $(PD\; +\; Df)\; -\; PF$, that is, less than $Df\; -\; DF$, or Aa. Again, let Qf, QF be lines drawn from a point within either of the hyperbolas, Qf must necessarily meet the curve; let D be the point of intersection, join FD; QF is less than $QD\; +\; DF$, and therefore $Qf\; -\; QF$ is greater than $Qf\; -\; (QD\; +\; DF)$, that is, greater than $Df\; -\; DF$ or Aa.

Cor. 2. A point is without or within the hyperbolas, according as the difference of two lines drawn from that point to the foci is less or greater than the transverse axis.

Cor. 3. The transverse axis is bisected in the centre. Hyperbola. Let C be the centre; then $CF\; =\; Cf$ (Def. 3), and $FA\; =\; fa$, therefore $CA\; =\; Ca$.

Let APB be a triangle, of which the side PA is greater than the side PB; draw a straight line from P, the vertex, to O, the middle of the base AB, and straight lines AQ, BQ, to any point in PO; the line QA will be greater than the line QB; and the excess of PA above PB will be greater than the excess of QA above QB.

Draw AC perpendicular to PO, and BD parallel to it, meeting AC in D, and join QD, PD, this last line meeting QB in E.

The triangles AOP, BOP have $AO\; =\; BO$, PO common to both, and PA greater than PB, therefore the angle AOQ is greater than the angle BOQ (25, 1, E.); and hence again, in the triangles AOQ, BOQ, the line AQ will be greater than BQ (24, 1, E.).

And because CO is parallel to DB, and $AO\; =\; OB$, therefore $AC\; =\; CD$ (3, 6, E.). The triangles ACP, DCP have thus $AC\; =\; DC$, CP common, and the angle ACP equal to DCP, therefore $PA\; =\; PD$; and in the same way it appears that $QA\; =\; QD$.

And since PA is greater than PB, and QA than QB, therefore PD is greater than PB, and QD than QB.

Now it was shown that $DE\; -\; EB\; >\; AQ\; -\; QB$; much more then is $AP\; -\; PB\; >\; AQ\; -\; QB$.

Every transverse diameter of a hyperbola is bisected in the centre.

Let Pp be a transverse diameter; it is bisected in C; for if CP and Cp be unequal, take CQ equal to CP; from the points P, p, Q, draw straight lines to F and f the foci. The triangles PCF, QCF have PC equal to QC and CF

Hyperbola equal to $Cf$ (Def. 3), and the angles at $C$ equal, therefore they are in every way equal, and $Pf$ is equal to $Qf$. In the same way, it appears that the triangles $PCf$, $QCf$ are equal, and that $Pf$ is equal to $Qf$; therefore $Pf\; -\; Pf\; =\; Qf\; -\; Qf$; but because $P$ and $q$ are points in the hyperbola $Pf\; -\; Pf\; =\; pf\; -\; pf$, therefore $pf\; -\; pf\; =\; Qf\; -\; Qf$. But by the preceding Lemma $pf\; -\; pf\; >\; Qf\; -\; Qf$; now this is impossible; therefore $CP$ and $Cp$ are not unequal, that is, they are equal.

COR. 1. Every transverse diameter meets the hyperbola in two points only.

COR. 2. Every transverse diameter divides the opposite hyperbolas into two parts which are equal and similar, the like parts being at opposite parts of the diameter.

The square of half the conjugate axis of an hyperbola is equal to the rectangle contained by the straight lines between either focus and the extremities of the transverse axis.

Draw a straight line from $A$, either of the extremities of the transverse axis, to $B$, either extremity of the conjugate axis. Then, $BC^2\; +\; CA^2\; =\; BA^2$ (47, 1, E.) $=\; CF^2$ (Def. 7). But because $Aa$ is bisected at $C$ and produced to $F$,

Let $ABC$ be a triangle, of which the side $BA$ is greater than the side $BC$; draw $BE$ perpendicular to the side $AC$, and straight lines $AD$, $CD$ to any point in $BE$; the line $DA$ will be greater than the line $DC$; and the excess of $DA$ above $DC$ will be greater than the excess of $BA$ above $BC$.

First, let the perpendicular $BE$ fall without the triangle $ABC$, and let $AD$ meet $BC$ in $H$.

The angle $DCA$ is greater than $BCA$; but $BCA$ being an obtuse angle, is greater than $BAC$, which is acute; and again $BAC$ is greater than $DAC$; much more then is $DCA$ greater than $DAC$; therefore $DA$ is greater than $DC$ (19, 1, E.).

And since $AH\; +\; HB\; >\; AB$,
therefore $AH\; +\; HB\; -\; BC\; >\; AB\; -\; BC$;
that is, $AH\; -\; HC\; >\; AB\; -\; BC$.

Again, since ,
therefore $AD\; -\; CD\; >\; AD\; -\; DH\; -\; CH$;
that is, $AD\; -\; CD\; >\; AH\; -\; CH$.

But it was shown that $AH\; -\; CH\; >\; AB\; -\; BC$;
much more then is $AD\; -\; CD\; >\; AB\; -\; BC$.

Next let the perpendicular fall within the triangle $ABC$; take $EC\; =\; FC$, and join $BC$, $DC$: then $BC\; =\; DC$,
and $DC\; =\; DC$,

and $BA\; -\; BC\; =\; BA\; -\; DC$,
also $DA\; -\; DC\; =\; DA\; -\; DC$;
but $DA\; -\; DC\; >\; BA\; -\; DC$,
therefore $DA\; -\; DC\; >\; BA\; -\; DC$.

To find any number of points in an hyperbola, having given the transverse axis and foci.

Let $F$, $f$ be the foci, $Aa$ the transverse axis, and $C$ the centre. Suppose the problem resolved, and that $D$ is a point in the hyperbola. Join $DF$, $Df$. Take $AH$ in the axis equal to $DF$; then $aH$ will be equal to $Df$ (Def. 1), and $HA\; +\; Ha\; =\; DF\; +\; Df$;
but $HA\; +\; Ha\; =\; HC\; +\; Ca\; +\; Ha\; =\; 2CH$,
therefore $DF\; +\; Df\; =\; 2CH$.

Now $DF\; +\; Df\; >\; Ff$,
therefore $2CH\; >\; Ff$ and $CH\; >\; Cf$.
Thus it appears that the point $H$ cannot be between the foci $Ff$, and that it may be anywhere in the line $Ff$, produced both ways.

CONSTRUCTION.—Take $H$, any point in the axis produced both ways, except between $F$ and $f$ the foci, and from $F$ and $f$ as centres, with the distances $HA$, $Ha$, describe circles which will cut each other in two points $Dd$, one on each side of the axis. These are points in the hyperbola.

Join $DF$, $Df$, also $dF$, $df$. Because $DF\; -\; Df\; =\; HA\; -\; Ha\; =\; Aa$, therefore $D$ is a point in the hyperbola; and in like manner it appears that $d$ is a point in the hyperbola.

In this way may any number of points be found.

COR. 1. Any perpendicular to the transverse axis which meets it produced either way, will cut the curve in two points, and in no more. For if the perpendicular $Dd$ could meet the curve in two points $D$, $D\text{'}$, on the same side of the axis, then $DF$, $Df$, also $D\text{'}F$, $D\text{'}f$, being drawn to the foci, $DF\; -\; Df$ would be equal to $D\text{'}F\; -\; D\text{'}f$, which is impossible (Lemma 2).

COR. 2. Every chord $Dd$ in an hyperbola, perpendicular to the transverse axis, is bisected by that axis, and therefore is an ordinate to it.

COR. 3. Of all the straight lines which can be drawn from either focus to either of the opposite hyperbolas, the shortest is that which passes through the centre (being produced if necessary); and only two equal straight lines can be drawn from either focus to one of the opposite hyperbolas, viz. one line on each side of the centre.

SCHOLIUM. From this proposition it appears that the opposite hyperbolas recede continually from the foci and from the axis, and that they are entirely separated from

each other, their nearest approach being at the vertices of the transverse axis. Also, that if the space Dad, bounded by either, were resolved into two by cutting it along the axis CH, the portions on each side of the axis would entirely coincide if one were turned over on the other. Now it was shown that the opposite hyperbolas might be applied one upon the other, viz. the curve Dad on $\backslash Delta\backslash Delta(2)$, therefore the transverse axis divides the opposite hyperbolas into four spaces, indefinite in extent, but which are exactly alike, and may be placed one on another, so as entirely to coincide.

The straight line which bisects the angle contained by two straight lines drawn from any point in the hyperbola to the foci is a tangent to the curve at that point.

Let D be the point in the curve, let DF, Df be straight lines drawn to the foci; the straight line DE which bisects the angle fDF, is a tangent to the curve.

Take H any other point in DE, take DG = Df, and join Hf, HG, fG; let fG meet DE in L. Because Df = DG, and DL is common to the triangles DfL, DGL, and the angles fDL, GLD are equal, these triangles are equal, and fL = LG, and hence fH = HG (4, 1, E.); and FH = fH = FH = HG; but since FH is less than FG + GH, FH = HG is less than FG, that is, less than FD = fD or Aa, therefore FH = fH is less than Aa; hence the point H is without the hyperbola (2 Cor. 1), and consequently DHL is a tangent to the curve at D (Def. 10).

COR. 1. There cannot be more than one tangent to the hyperbola at the same point. For D is such a point in the line DE, that the difference of the lines DF, Df, the distances of that point from the foci, is evidently greater than the difference of FH, fH, the distances of H, any other point in that line; and if another line KD be drawn through D, there is in like manner a point K in that line, which will be different from D, such that the difference of FK, fK is greater than the difference of the distances of any other point in KD, and therefore greater than FD = fD; therefore the point K will be within the hyperbola (2 Cor. 1), and the line KD will cut the curve.

COR. 2. A perpendicular to the transverse axis at either of its extremities is a tangent to the curve. The demonstration is the same as for the proposition, if it be considered that when D falls at either extremity of the axis, the point L falls also at the extremity of the axis, and thus the tangent DE, which is always perpendicular to fL, is perpendicular to the axis.

COR. 3. Every tangent to either of the opposite hyperbolas passes between that hyperbola and the centre. Let the tangent DL meet the axis in E. Because DE bisects the angle fDF,

But FD is greater than fD (Def. 1), therefore FE is Hyperbola greater than fE, and hence E is between C and the vertex of the hyperbola to which DE is a tangent.

SCHOLIUM. From the property of the hyperbola which forms this proposition, the points F and f are called foci; for rays of light proceeding from one focus, and falling upon a polished surface whose figure is that formed by the revolution of the curve about the transverse axis, are reflected in lines passing through the other focus.

The tangents at the vertices of any transverse diameter of an hyperbola are parallel.

Let Pp be a diameter, HP, hp tangents at its vertices; draw straight lines from P and p to F and f, the foci. The triangles FCP, fCp, having FC = fC, CP = Cp (2), and the angles at C equal, are in all respects equal; and because the angle FPC is equal to Cpf, FP is parallel to fp (27, 1, E.), therefore Pf is equal and parallel to pF (33, 1, E.); thus FPfp is a parallelogram of which the opposite angles P and p are equal (34, 1, E.). Now the angles FPH, spH are the halves of these angles (4), therefore the angles FPH, spH, and hence CPH, Cph, are also equal, and consequently HP is parallel to hp.

COR. 1. If tangents be drawn to an hyperbola at the vertices of a transverse diameter, straight lines drawn from either focus to the points of contact make equal angles with these tangents; for the angle Fph is equal to FPH.

COR. 2. The transverse axis is the only diameter which is perpendicular to tangents at its vertices. For let Pp be any other diameter. The angle CPH is less than FPH, that is, less than the half of FPf, therefore CPH is less than a right angle.

If a straight line be drawn from either focus of an hyperbola to the intersection of two tangents to the curve, it will make equal angles with straight lines drawn from the same focus to the point of contact.

Hyperbola. Let $HP,\; Hp$, tangents to an hyperbola at $P$ and $p$, intersect each other at $H$; draw $PF,\; pF,\; HF$, to $F$, either of the foci; the line $HF$ makes equal angles with $FP,\; Fp$. Draw $Pf,\; pf$ to $f$, the other focus, and in $FP,\; Fp$ take $PK\; =\; Pf$ and $pk\; =\; pf$; join $HK,\; Hk$, and draw $fK,\; fk$, meeting the tangents in $G$ and $g$.

The triangles $HPK,\; HPf$ have $PK\; =\; Pf,\; PH$ common to both, and the angles $KPH,\; fPH$ equal (5); therefore they are in every way equal, and have $HK\; =\; Hf$. In the same way it may be shown, that the triangles $Hpk,\; Hpf$ are in every way equal, and therefore that $Hk\; =\; Hf$.

The triangles $HPK,\; HPf$ have $HK\; =\; Hk$ (for each is equal to $Hf$), $HF$ common to both, and $FK\; =\; Fk$, because each is equal to $PF\; =\; Pf$, or $pF\; =\; pf$, that is, to the transverse axis; therefore they are in all respects equal; and the angle $HPK$ is equal to the angle $HPf$; therefore $HF$ makes equal angles with $FP$ and $Fp$.

Cor. Perpendiculars drawn from the intersection of two tangents to straight lines drawn from either focus through the points of contact are equal. Let $HI,\; Hi$ be perpendiculars drawn from $H$, the intersection of the tangents $PH,\; pH$ on the lines $FP,\; Fp$. The triangles $HFI,\; HFi$ are in all respects equal (26, 1, E.); therefore $HI\; =\; Hi$.

Straight lines drawn from the intersection of two tangents to the foci, make equal angles with the tangents.

Let $F,\; f$ be the foci of an hyperbola, and let straight lines $HP,\; Hp$, which intersect each other at $H$, touch the hyperbola at $P$ and $p$; also let $HF,\; Hf$ be lines drawn to the foci; and let $FH$ be produced to any distance $O$, the angles $PHO,\; pHf$ are equal.

The same construction being made as in Prop. 7, because the angles $FHK,\; FHK$ are equal, the angles $KHO,\; kHO$ are equal.

therefore $2PHO\; =\; 2pHf$, and $PHO\; =\; pHf$.

If two tangents to an hyperbola be at the extremities of a chord, and a third tangent be parallel to the chord, the part of this tangent intercepted by the other two is bisected at the point of contact.

Let $HD,\; Hd$ be tangents at the extremities of the chord $Dd$; and let $Kk$, a tangent parallel to $Dd$, meet the other two tangents in $K$ and $k$. The intercepted segment $Kk$ is bisected at $P$, the point of contact.

From the points of contact $D,\; P,\; d$, draw lines to $F$, either of the foci; and from $H,\; K,\; k$, the intersections of the tangents, draw perpendiculars to the lines drawn from the points of contact to the foci, viz. $HI,\; Hi$ perpendicular to $DF,\; dF$, and $KM,\; KN$ perpendicular to $ED,\; FP$, and $km,\; kn$ perpendicular to $Fd,\; FP$.

The triangles $DHI,\; DKM$ are manifestly equiangular, also the triangles $dHi,\; dkm$;

But because $Dd$ is parallel to $Kk$, a side of the triangle $HKk$,

Now $HI\; =\; Hi$ (Cor. 7), therefore $KM\; =\; km$; but $KM\; =\; KN$, and $km\; =\; kn$ (Cor. 7), therefore $KN\; =\; kn$; and since from the similar triangles $KPN,\; kPn$, $KN\; :\; kn\; =\; KP\; :\; kP$, therefore $KP$ is equal to $kP$.

Let $KLl$ be a triangle, having its base $Ll$ bisected at $p$, and let $Hh$, any straight line parallel to the base, and terminated by the sides produced, be bisected at $P$, then $P,\; p$, the points of bisection, and $K$, the vertex of the triangle, are in the same straight line, and that line bisects $Dd$ any other line parallel to the base.

Join $KP,\; Kp$; the triangles $KHh,\; KLl$ being similar, and $Hh,\; Ll$ similarly divided at $P,\; p$,

Now the angles at $H$ and $L$ are equal, therefore the triangles $KHP,\; KLp$ are similar, and the angle $PKH$ is equal to $pKL$; to both add the angle $HKp$, and the angles $PKH,\; HKp$ are equal to $pKL,\; HKp$, that is, to two right angles; therefore $KP,\; Kp$ lie in the same straight line (14, 1, E).

Next let $Dd$ meet $KP$ in $E$, then

Any chord not passing through the centre, but parallel to a tangent, is bisected by the diameter which passes

through the point of contact; or it is an ordinate to that diameter.

The chord $DEd$, which is parallel to $Kk$, a tangent at $P$, is bisected at $E$ by the diameter $PCp$.

Draw $Lp$, a tangent at $p$, the other end of the diameter, and $DH$, $dH$ tangents at $D$ and $d$, the extremities of the chord meeting the other tangents in $K$, $k$ and $L$, $l$. Then $KPa$ and $Lpl$ are bisected at $P$ and $p$ (9), therefore the diameter $Pp$, when produced, will pass through $H$, and bisect $Dd$, which is parallel to $Kk$ or $Ll$ in $E$. (Lemma.)

Cor. 1. Straight lines which touch an hyperbola in the extremities of an ordinate to any diameter, intersect each other in that diameter.

Cor. 2. Every ordinate to a diameter is parallel to a tangent at its vertex; for if not, let a tangent be drawn parallel to the ordinate: then the diameter drawn through the point of contact would bisect the ordinate, and thus the same diameter would be bisected in two different points, which is absurd.

Cor. 3. All the ordinates to the same diameter are parallel to each other.

Cor. 4. A straight line that bisects two parallel chords, and terminates in the curve, is a diameter.

Cor. 5. The ordinates to either axis are perpendicular to that axis, and no other diameter is perpendicular to its ordinates.

PROP. XI.

If a tangent to an hyperbola meet a transverse diameter, and from the point of contact an ordinate be drawn to that diameter, the semidiameter will be a mean proportional between the segments of the diameter intercepted between the centre and the ordinate, and between the centre and the tangent.

(See Figure to Prop. X.)

Let $DH$, a tangent to the hyperbola at $D$, meet a transverse diameter $Pp$ produced in $H$, and let $DEd$ be an ordinate to that diameter: $CE\; :\; CP\; =\; CP\; :\; CH$.

Through $P$ and $p$, the vertices of the diameter, draw the tangents $PK$, $pL$, meeting $DH$ in $K$ and $L$: draw $PF$, $pF$ to either of the foci, and draw $KM$ and $KN$ perpendicular to $FD$ and $FP$, and also $LO$ and $LI$ perpendicular to $FD$ and $Fp$.

The triangles $PKN$, $pLI$ are equiangular, for the angles at $N$ and $I$ are right angles, and the angles $NPK$, $IpL$ are equal (1 Cor. 6); therefore

$PK\; :\; pL\; =\; KN\; :\; LI$ (4, 6, E.) $=\; KM\; :\; LO$ (Cor. 7).

But the triangles $KDM$, $LDO$ being manifestly equiangular, $KM\; :\; LO\; =\; KD\; :\; LD$; therefore $PK\; :\; pL\; =\; KD\; :\; LD$.

But because of the parallel lines $PK$, $ED$, $pL$, the triangles $HPK$, $HpL$ are equiangular, and the lines $HL$, $Hp$ are similarly divided in $K$, $D$, and in $P$, $E$ (10, 6, E.); hence $PK\; :\; pL\; =\; HP\; :\; Hp$, and $KD\; :\; LD\; =\; PE\; :\; pE$; therefore $HP\; :\; Hp\; =\; PE\; :\; pE$.

Take $CG\; =\; CE$, and then $PE\; =\; pG$, and, by composition, $HP\; :\; Pp\; =\; PE\; :\; EG$; and taking the halves of the consequents, $HP\; :\; PC\; =\; PE\; :\; EC$, and, by division, $HC\; :\; PC\; =\; PC\; :\; EC$.

Cor. 1. The rectangle $PE\; \backslash cdot\; Ep$ is equal to the rectangle $HE\; \backslash cdot\; EC$.

For $PC^2\; =\; HC\; \backslash cdot\; CE$ (17, 6, E.) $=\; EC^2\; -\; HE\; \backslash cdot\; EC$ (8, 2, E.), also $PC^2\; =\; EC^2\; -\; PE\; \backslash cdot\; Ep$ (5, 2, E.); therefore $HE\; \backslash cdot\; EC\; =\; PE\; \backslash cdot\; Ep$.

Cor. 2. The rectangle $PH\; \backslash cdot\; Hp$ is equal to the rectangle $HE\; \backslash cdot\; HC$.

For $HC^2\; =\; CP^2\; -\; PH\; \backslash cdot\; Hp$ (6, 2, E.), and $HC^2\; =\; EC\; \backslash cdot\; HC\; -\; EH\; \backslash cdot\; HC$ (1, 2, E.) $=\; CP^2\; -\; EH\; \backslash cdot\; HC$ (by the Prop.); therefore $PH\; \backslash cdot\; Hp\; =\; EH\; \backslash cdot\; HC$.

PROP. XII.

In $Aa$, the transverse axis of an hyperbola, let there be taken on each side of the centre straight lines $CK$, $Ck$, each a fourth proportional to $CF$ the eccentricity, and $CA$, $CB$, half the transverse and conjugate axes: if then from $P$, a vertex of any diameter, there be drawn $PH$ perpendicular to $Aa$; the square of the semidiameter $PC$, will have to the rectangle contained by the segments $KH$, $kH$ the constant ratio of the square of $CF$ to the square of $CA$.

Draw $PL$ perpendicular to the conjugate axis. Because $CB^2\; :\; CA^2\; =\; CB^2\; +\; CL^2\; :\; PL^2$, by composition, $CB^2\; +\; CA^2\; :\; CA^2\; =\; CB^2\; +\; CL^2\; +\; PL^2\; :\; PL^2$, therefore (47, 1, E. and Def. 8) $CF^2\; :\; CA^2\; =\; CB^2\; +\; PC^2\; :\; PL^2$ or $CH^2$;

but, by hypothesis, $CF^2\; :\; CA^2\; =\; CB^2\; :\; CK^2$, therefore $CB^2\; :\; CK^2\; =\; CB^2\; +\; PC^2\; :\; CH^2$; and hence (19, 5, E.) $PC^2\; :\; CH^2\; =\; CK^2\; :\; CA^2$, and (6, 2, E.) $PC^2\; :\; KH\; \backslash cdot\; Hk\; =\; CF^2\; :\; CA^2$.

Cor. 1. Hence the squares of any semidiameters $PC$, $QC$ are to one another as the rectangles $KH\; \backslash cdot\; Hk$, $KI\; \backslash cdot\; Ik$, contained by the segments of the line $Kk$ between its extremities, and perpendiculars from the vertices of the diameters.

Cor. 2. The transverse axis is the least of all the diameters, and a diameter which is nearer to the transverse axis is less than one more remote, and a semidiameter may be found greater than any given line.

By hypothesis $CF$ or $AB\; :\; CA\; =\; CB\; :\; CK$: Now $AB$ is greater than $CB$ (19, 1, E.), therefore $CA$ is greater than $CK$, and the points $K$, $k$ are between $Aa$, the vertices of the transverse axis. Suppose now a semidiameter $PC$ to

Hyperbola turn about C, and that in every position PH is perpendicular to KA produced both ways; the rectangle KH · HA, and the square of PC, to which the rectangle has a constant ratio, will manifestly be least when PC coincides with AC, and both will increase as H recedes from C; and, as the rectangle may exceed any given space, the semidiameter may become greater than any given line.

Cor. 3. Diameters which make equal angles with the transverse axis on opposite sides of it are equal; and only two equal diameters can be drawn, one on each side of the transverse axis.

If a tangent to an hyperbola meet the conjugate axis, and from the point of contact a perpendicular be drawn to that axis, the semiaxis will be a mean proportional between the segments of the axis intercepted between the centre and the perpendicular, and between the centre and the tangent.

Let DH, a tangent to the hyperbola at D, meet the conjugate axis $Bb$ in H, and let DG be perpendicular to that axis, then $CG\; :\; CB\; =\; CB\; :\; CH$.

Let DH meet the transverse axis in K; draw DE perpendicular to that axis, draw DF, $Df$ to the foci, and describe a circle about the triangle $DfF$; the conjugate axis will evidently pass through the centre of the circle; and because the angle $DFf$ is bisected by the tangent DK, the line DK will pass through one extremity of the diameter; therefore the circle passes through H. Draw DL to the other extremity of the diameter. The triangles $LGD$, $KCH$ are similar, for each is similar to the right angled triangle $LDH$, therefore

hence $LG\; \backslash cdot\; CH\; =\; CE\; \backslash cdot\; CK\; =\; CA^2$ (11.)

Now $LC\; \backslash cdot\; CH\; =\; CF^2$ (35, 3, E.),

therefore $LC\; \backslash cdot\; CH\; -\; LG\; \backslash cdot\; CH\; =\; CF^2\; -\; CA^2$;

that is, $CG\; \backslash cdot\; CH\; =\; CB^2$ (Def. 7),

wherefore $CG\; :\; CB\; =\; CB\; :\; CH$.

XI. If through A, one of the vertices of the transverse axis, a straight line $HAh$ be drawn, equal and parallel to $Bb$ the conjugate axis, and bisected at A by the transverse axis; the straight lines $CHM$, $Chm$ drawn through the centre, and the extremities of that parallel, are called Asymptotes.

Cor. 1. The asymptotes of two opposite hyperbolas are common to both. Through $a$, the other extremity of the axis, draw $H\text{'}ah\text{'}$, parallel to $Bb$, and meeting the asymptotes of the hyperbola $DAD\text{'}$ in $H\text{'}$ and $h\text{'}$. Because $aC$ is equal to $AC$, $aH\text{'}$ is equal to $Ah$, and $ah\text{'}$ to $AH$.

Cor. 2. The asymptotes are diagonals of a rectangle formed by drawing perpendiculars to the axes at their vertices; for the lines $AH$, $CB$, $aH\text{'}$, being equal and parallel, the points $H$, $B$, $H\text{'}$ are in a straight line parallel to $Aa$; the same is true of the points $h$, $b$, $h\text{'}$.

The asymptotes do not meet the hyperbola; and if from any point in the curve a straight line be drawn parallel to the conjugate axis, and terminated by the asymptotes, the rectangle contained by its segments from that point is equal to the square of half the axis.

Through D, any point in the hyperbola, draw a straight line parallel to the conjugate axis, meeting the transverse axis in E, and the asymptotes in M and m; the points M and m shall be without the hyperbola, and the rectangle $MD\; \backslash cdot\; Dm$ equal to the square of $BC$.

Draw DG perpendicular to $Bb$ the conjugate axis; let a tangent to the curve at D meet the transverse axis in K, and the conjugate axis in L, and let a perpendicular at the vertex A meet the asymptote in H. Because DK is a tangent, and DE an ordinate to the axis, CA is a mean proportional between CK and CE (11); and therefore

But $CK\; :\; CE\; =\; LC\; :\; LG$,

and $CA^2\; :\; CE^2\; ::\; AH^2\; :\; EM^2$,

therefore $LC\; :\; LG\; ::\; AH^2\; :\; EM^2$.

Again, CB being a mean proportional between CL and CG (13),

and therefore, by composition,

wherefore $AH^2\; :\; EM^2\; =\; CB^2\; :\; CB^2\; +\; ED^2$.

therefore $EM^2\; =\; CB^2\; +\; ED^2$,

consequently $EM^2$ is greater than $ED^2$, and $EM$ greater than $ED$, therefore M is without the hyperbola. In like manner it appears that m is without the hyperbola, therefore every point in both the asymptotes is without the hy-

perbola. Again, the straight line $Mm$, terminated by the asymptotes, being manifestly bisected by the axis at $E$,
$ME^2\; =\; MD\; \backslash cdot\; Dm\; +\; DE^2$;
but it has been shown that

therefore $MD\; \backslash cdot\; Dm\; =\; BC^2$.

COR. 1. Hence, if in a straight line $Mm$, terminated by the asymptotes, and parallel to the conjugate axis, there be taken a point $D$ such that the rectangle $MD\; \backslash cdot\; Dm$ is equal to the square of that axis, the point $D$ is in the hyperbola.

COR. 2. If straight lines $MDm,\; NRn$, be drawn through $D$ and $R$, any points in the hyperbola, or opposite hyperbolas, parallel to the conjugate axis, and meeting the asymptotes in $M,\; m$ and $N,\; n$, the rectangles $MD\; \backslash cdot\; Dm,\; NR\; \backslash cdot\; Rn$ are equal.

The hyperbola and its asymptote, when produced, continually approach to each other, and the distance between them becomes less than any given line.

Take two points $E$ and $O$ in the transverse axis produced, and through these points draw straight lines parallel to the conjugate axis, meeting the hyperbola in $D,\; R$, and the asymptotes in $M,\; m$ and $N,\; n$.

therefore the point $R$ is nearer to the asymptote than $D$, that is, the hyperbola when produced approaches to the asymptote.

Let $S$ be any line less than half the conjugate axis; then, because $Dm$, a straight line drawn from a point in the hyperbola, parallel to the conjugate axis, and terminated by the asymptote on the other side of the transverse axis, may evidently be of any magnitude greater than $AS$, which is equal to half the conjugate axis, $Dm$ may be a third proportional to $S$ and $BC$; and since $Dm$ is also a third proportional to $DM$ (the segment between $D$ and the other asymptote) and $BC$, $DM$ may be equal to $S$; but the distance of $D$ from the asymptote is less than $DM$, therefore that distance may become less than $S$, and consequently less than any given line.

COR. Every straight line passing through the centre within the angles contained by the asymptotes through which the transverse axis passes, meets the hyperbola,

and therefore is a transverse diameter; and every straight line passing through the centre within the adjacent angles falls entirely without the hyperbola.

SCHOLIUM. The name asymptotes (non concurrentes) has been given to the lines $GH,\; CH$, because of the property they have of continually approaching to the hyperbola without meeting it, as has been proved in this proposition.

If from two points in an hyperbola, or opposite hyperbolas, two parallel straight lines be drawn to meet the asymptotes, the rectangles contained by their segments between the points and the asymptotes are equal.

Let $D$ and $G$ be two points in the hyperbola, or opposite hyperbolas: let parallel lines $EDe,\; HGH$ be drawn to meet the asymptotes in $E,\; e$ and $H,\; h$; the rectangles $ED\; \backslash cdot\; De,\; HG\; \backslash cdot\; Gh$ are equal.

Through $D$ and $G$ draw straight lines parallel to the conjugate axis, meeting the asymptotes in the points $L,\; l$ and $M,\; m$. The triangles $HGM,\; EDL$ are similar, as also the triangles $Hgm,\; eDl$.

hence, taking the rectangles of the corresponding terms of the proportions,

COR. 1. If a straight line be drawn through $D,\; d$, two points in the same, or opposite hyperbolas, the segments $DE,\; de$ between those points and the asymptotes are equal. For in the same manner that the rectangles $ED\; \backslash cdot\; De,\; GH\; \backslash cdot\; HG$ have been proved to be equal, it may be shown that the rectangles $Ed\; \backslash cdot\; de,\; HG\; \backslash cdot\; Gh$ are equal, therefore $ED\; \backslash cdot\; De\; =\; Ed\; \backslash cdot\; de$. Let $Ee$ be bisected in $O$, then $ED\; \backslash cdot\; De\; =\; EO^2\; -\; OD^2$, and $Ed\; \backslash cdot\; de\; =\; EO^2\; -\; Od^2$, therefore $EO^2\; -\; OD^2\; =\; EO^2\; -\; Od^2$; hence $OD\; =\; Od$, and $ED\; =\; ed$.

COR. 2. When the points D and d are in the same hyperbola, by supposing them to approach till they coincide at P, the line $De$ will thus become a tangent to the curve at P. Therefore any tangent $KPk$, which is terminated by the asymptotes, is bisected at P, the point of contact.

COR. 3. And if any straight line $KPk$, limited by the asymptotes, be bisected at P, a point in the curve, that line is a tangent at P. For it is evident that only one line can be drawn through P, which shall be limited by the asymptotes, and bisected at P.

COR. 4. If a straight line be drawn through D, any point in the hyperbola, parallel to a tangent $KPk$, and terminated by the asymptotes at E and e, the rectangle $ED\; \backslash cdot\; De$ is equal to the square of $PK$, the segment of the tangent between the point of contact and either asymptote. The demonstration is the same as in the proposition.

COR. 5. If from any point D in a hyperbola a straight line be drawn parallel to $Pp$, any diameter, meeting the asymptotes in E and e, the rectangle $ED\; \backslash cdot\; De$ is equal to the square of half the diameter. The demonstration is the same as in the proposition.

If two straight lines be drawn from any point in a hyperbola to the asymptotes, and from any other point in the same or opposite hyperbolas two other lines be drawn parallel to the former; the rectangle contained by the first two lines will be equal to the rectangle contained by the other two lines.

From D, any point in the hyperbola, draw $DH$ and $DK$ to the asymptotes, and from any other point $d$ draw $dh$ and $dk$ parallel to $DH$ and $DK$. The rectangles $HD\; \backslash cdot\; DK$, $hd\; \backslash cdot\; dk$ are equal.

Join $D,\; d$, cutting the asymptotes in $E,\; e$. From similar triangles

therefore, taking the rectangles of corresponding terms,

COR. 1. If the lines $DK\text{'},\; D\text{'}H\text{'},\; d\text{'}k\text{'},\; dk\text{'}$ be parallel to the asymptotes, and thus form the parallelograms $DK\text{'}CH\text{'},\; d\text{'}k\text{'}Ch\text{'}$, these are equal to one another (16, and 14, 6, E.). And if $D\text{'}C,\; d\text{'}C$ be joined, the halves of the parallelograms, or the triangles $D\text{'}KC,\; d\text{'}k\text{'}C$ are also equal.

COR. 2. If from $D\text{'},\; d\text{'}$, any two points in an hyperbola, straight lines $DK\text{'},\; d\text{'}k\text{'}$ be drawn parallel to one asymptote, meeting the other in $K\text{'}$ and $k\text{'}$, these lines are to each other reciprocally as their distances from the centre, or $D\text{'}K\text{'}\; :\; d\text{'}k\text{'}\; ::\; CK\text{'}\; :\; Ck\text{'}$. This appears from last corollary, and 14, 6, E.

XII. If $Aa$ be the transverse axis and $Bb$ the conjugate axis of two opposite hyperbolas $DAD,\; dad$, and if $Bb$ be the transverse axis and $Aa$ the conjugate axis of other two opposite hyperbolas $EBE,\; ebe$, these hyperbolas are said to be conjugate to the former. When all the four hyperbolas are mentioned they are called conjugate hyperbolas.

COR. The asymptotes of the hyperbolas $DAD,\; dad$ are also the asymptotes of the hyperbolas $EBE,\; ebe$. This is evident from Cor. 2 to Definition 11.

XIII. Any diameter of the conjugate hyperbolas is called a second diameter of the other hyperbolas.

COR. Every straight line passing through the centre, within the angle through which the conjugate or second axis passes, is a second diameter of the hyperbola.

XIV. Any straight line not passing through the centre, but terminated both ways by the opposite hyperbolas, and bisected by a second diameter, is called an ordinate to that diameter.

Any straight line not passing through the centre, but terminated by the opposite hyperbolas, and parallel to a tangent to either of the conjugate hyperbolas, is bisected by the second diameter that passes through the point of contact, or is an ordinate to that diameter.

The straight line $Dd$ terminated by the opposite hyperbolas, and parallel to the tangent $KQk$, is bisected at $E$ by $Qq$, the diameter that passes through the point of contact.

Let $Dd$ meet the asymptotes in $G$ and $g$, and let the tangent meet them in $K$ and $k$. The straight lines $Gg,\; Kk$ are evidently similarly divided at $E$ and $Q$, and since $KQ\; =\; Qk$ (2 Cor. 16), therefore $GE\; =\; Eg$; now $DG\; =\; gd$ (1 Cor. 16), therefore $DE\; =\; Ed$.

COR. 1. Every ordinate to a second diameter is parallel to a tangent at its vertex. The demonstration is the same as in Cor. 2, Prop. 10.

COR. 2. All the ordinates to the same second diameter are parallel to each other.

COR. 3. A straight line that bisects two parallel straight lines which terminate in the opposite hyperbolas is a second diameter.

COR. 4. The ordinates to the conjugate or second axis are perpendicular to it, and no other second diameter is perpendicular to its ordinates.

COR. 5. The opposite hyperbolas are similar to one another, and like portions of them are in all respects equal.

If a transverse diameter of an hyperbola be parallel to the ordinates to a second diameter; the latter shall be parallel to the ordinates to the former.

Let Pp, a transverse diameter of an hyperbola, be parallel to Dd, any ordinate to the second diameter Qq; the second diameter Qq shall be parallel to the ordinates to the diameter Pp.

Draw the diameter dd through one extremity of the ordinate dD, and join d and D, the other extremity, meeting Pp in E. Because dd is bisected at C, and CE is parallel to dD, the line Dd is bisected at E, therefore Dd is an ordinate to the diameter Pp. And because Dd and dd are bisected at G and C, the diameter Qq is parallel to Dd (2, 6, E.); therefore Qq is parallel to any ordinate to the diameter Pp.

XV. Two diameters are said to be conjugate to one another when each is parallel to the ordinates to the other diameter.

COR. Diameters which are conjugate to one another are parallel to tangents at the vertices of each other.

XVI. A third proportional to any diameter and its conjugate is called the Parameter, also the Latus rectum of that diameter.

The tangent at the vertex of any transverse diameter of an hyperbola, which is terminated by the asymptotes, is equal to the diameter that is conjugate to that diameter.

Let Pp be any transverse diameter of an hyperbola, HPa a tangent at its vertex, meeting the asymptotes in H and h, and Qq the diameter which is conjugate to Pp; the tangent Hh is equal to the diameter Qq.

Through D, any point in the hyperbola, draw a straight line parallel to the tangent and diameter, cutting either of the conjugate hyperbolas in d, and the asymptotes in I and i, and through D and d draw lines parallel to Bb, the conjugate axis, meeting the asymptotes in the points K, k, and L, l. The triangles DIK, dIL are similar, as also iDk, idl, therefore

therefore, taking the rectangles of the corresponding terms,

But $KD\; \backslash cdot\; Dk\; =\; BC^2$ (14), and $Id\; \backslash cdot\; dl\; =\; Ld\; \backslash cdot\; dl$ (5 Cor. 16), therefore $ID\; \backslash cdot\; Di\; =\; Id\; \backslash cdot\; di$.

COR. 1. If another tangent be drawn to the curve at p, meeting the asymptotes in H' and h', the straight lines which join the points H, H', also h, h', are tangents to the conjugate hyperbolas at Q and q. For pH', as well as PH, is equal and parallel to CQ; therefore the points H, Q, H', are in a straight line parallel to Pp, and HQ = H'Q (33, 1, E.), therefore HQH is a tangent to the curve at Q. In like manner it appears that hqh' is a tangent at q.

COR. 2. If tangents be drawn at the vertices of two conjugate diameters, they will meet in the asymptotes, and form a parallelogram, of which the asymptotes are diagonals.

If a tangent to an hyperbola meet a second diameter, and from the point of contact an ordinate be drawn to that diameter; half the second diameter will be a mean proportional between the segments of the diameter intercepted between the centre and the ordinate and between the centre and the tangent.

Let DL, a tangent to the curve at D, meet the second diameter Qq in L, and let DGe be an ordinate to that diameter; then

Let Pp be the diameter that is conjugate to Qq, let HPa be a tangent at the vertex, terminated by the asymptotes; through D draw the ordinate DEd to the diameter Pp, meeting the asymptotes in M and m; let K be the intersection of DL and Pp. Because DK is a tangent at D, and DEd an ordinate to Pp, CP is a mean proportional between CE and CK (11), and therefore

Now, the lines CQ, PH, EM, being parallel (10, and Def. 15), from similar triangles,

If an ordinate be drawn to any transverse diameter of an hyperbola; the rectangle under the abscisses of the diameter will be to the square of the semi-ordinate as the square of the diameter to the square of its conjugate.

Let $DEd$ be an ordinate to the transverse diameter $Pp$, and let $Qq$ be its conjugate diameter;

Let $DKL$, a tangent at $D$, meet the diameter in $K$, and its conjugate in $L$. Draw $DG$ parallel to $Pp$, meeting $Qq$ in $G$. Because $CP$ is a mean proportional between $CE$ and $CK$ (11),

and because $CQ$ is a mean proportional between $CG$ and $CL$ (21),

COR. 1. If an ordinate be drawn to any second diameter of an hyperbola, the sum of the squares of half the second diameter and its segment, intercepted by the ordinate from the centre, is to the square of the semi-ordinate as the square of the second diameter to the square of its conjugate.

Let $DG$ be a semi-ordinate to the second diameter $Qq$. It has been shown that

COR. 2. The squares of semi-ordinates, and of ordinates to any transverse diameter, are to one another as the rectangles contained by the corresponding abscisses; and the squares of semi-ordinates, and of ordinates to any second diameter, are to one another as the sums of the squares of half that diameter, and the segments intercepted by the ordinate from the centre.

COR. 3. The ordinates to any transverse diameter, which intercept equal segments of that diameter from the centre, are equal to one another, and, conversely, equal ordinates intercept equal segments of the diameter from the centre.

The transverse axis of an hyperbola is the least of all its transverse diameters, and the conjugate axis is the least of all its second diameters.

Let $Rr$ be the transverse axis, $Pp$ any other transverse diameter; draw $PE$ perpendicular to $Rr$; then $CE$ being greater than $CR$, and $CP$ greater than $CE$, much more is $CP$ greater than $CR$, therefore $Pp$ is greater than $Rr$. In like manner it is shown, that if $Ss$ be the conjugate axis, and $Qq$ any other second diameter, $Qq$ is greater than $Ss$.

If an ordinate be drawn to any transverse diameter of an hyperbola, the rectangle under the abscisses of the diameter is to the square of the semi-ordinate as the diameter to its parameter.

Let $DE$ be a semi-ordinate to the transverse diameter $Pp$; let $PG$ be the parameter of the diameter, and $Qq$ the conjugate diameter. By the definition of the parameter (Def. 16),

COR. Let the parameter $PG$ be perpendicular to the diameter $Pp$; join $pG$, and from $E$ draw $EM$ parallel to $PG$, meeting $pG$ in $M$. The square of $DE$, the semi-ordinate, is equal to the rectangle contained by $PE$ and $EM$.

If the rectangles $PGLp$, $HGKM$ be completed, it will appear that the square of $ED$ is equal to the rectangle $MP$, which rectangle is greater than the rectangle $KP$, contained by the absciss $PE$, and the parameter $GP$, by a rectangle $KH$ similar and similarly situated to $LP$, the rectangle contained by the parameter and diameter. It

was on account of the excess of the ordinate above the rectangle contained by the absciss and parameter that Apollonius gave the curve to which the property belonged the name of Hyperbola.

If from the vertices of two conjugate diameters of an hyperbola there be drawn ordinates to any third transverse diameter; the square of the segment of that diameter, intercepted between the ordinate from the vertex of the second diameter and the centre, is equal to the rectangle contained by the segments between the other ordinate and the vertices of the third transverse diameter. And the square of the segment intercepted between the ordinate from the vertex of the transverse diameter and the centre is equal to the square of the segment between the other ordinate and the centre, together with the square of half the third transverse diameter.

Let $Pp,\; Qq$ be two conjugate diameters, of which $Pp$ is a transverse and $Qq$ a second diameter; let $PE,\; QG$ be semi-ordinates to any third transverse diameter $Rr$; then $CG^2\; =\; RE\; \backslash cdot\; Er$, and $CE^2\; =\; CG^2\; +\; CR^2$.

Draw the tangents $PH,\; QK$, meeting $Rr$ in $H$ and $K$. The rectangles $HC\; \backslash cdot\; CE$ and $CK\; \backslash cdot\; CG$ are equal; for each is equal to $CR^2$ (11), therefore

But the triangles $HPC,\; CQK$ are evidently similar (Cor. Def. 15); and since $PE,\; QG$ are parallel, their bases $CH,\; KC$ are similarly divided at $E$ and $G$, therefore

wherefore $CG\; :\; CE\; =\; HE\; :\; CG$, consequently $CG^2\; =\; CE\; \backslash cdot\; EH\; =\; (1\; \backslash text\{\; Cor.\; 11\})\; RE\; \backslash cdot\; Er$. Again, from the similar triangles $HPC,\; CQK$, $HC\; :\; CK\; =\; CE\; :\; KG$.

Now, it was shown that $HC\; :\; CK\; =\; CG\; :\; CE$, therefore $CG\; :\; CE\; =\; CE\; :\; KG$, consequently

but $GC\; \backslash cdot\; CK\; =\; CR^2$ (21), therefore $CE^2\; =\; CG^2\; +\; CR^2$.

Cor. 1. Let $Ss$ be the diameter that is conjugate to $Rr$, then $Rr$ is to $Ss$ as $CG$ to $PE$, or as $CE$ to $QG$.

For $Rr^2\; :\; Ss^2\; =\; RE\; \backslash cdot\; Er$, or $CG^2\; :\; PE^2$, therefore $Rr\; :\; Ss\; =\; CG\; :\; PE$. In like manner $Rr\; :\; Ss\; =\; CE\; :\; QG$.

Cor. 2. The difference between the squares of $CE,\; CG$, the segments of the transverse diameter to which the semi-ordinates $PE,\; QG$ are drawn, is equal to the square of $CR$ the semi-diameter. For it has been shown that

therefore $CE^2\; -\; CG^2\; =\; CR^2$. Cor. 3. The difference of the squares of any two conjugate diameters is equal to the difference of the squares of the axes. Let $Rr,\; Ss$ be the axes, and $Pp,\; Qq$ any two conjugate diameters; draw $PE,\; QG$ perpendicular to $Rr$, and $PL,\; QM$ perpendicular to $Ss$. Then

$CE^2\; -\; CG^2\; =\; CR^2$, and $CM^2\; -\; CL^2$, or $GQ^2\; -\; PE^2\; =\; CS^2$; therefore $CE^2\; +\; PE^2\; -\; (CG^2\; +\; GQ^2)\; =\; CR^2\; -\; CS^2$; that is (47, 1, E.), $CP^2\; -\; CQ^2\; =\; CR^2\; -\; CS^2$, therefore $Pp^2\; -\; Qq^2\; =\; Rr^2\; -\; Ss^2$.

If four straight lines be drawn touching conjugate hyperbolas at the vertices of any two conjugate diameters; the parallelogram formed by these lines is equal to the rectangle contained by the transverse and conjugate axes.

Let $Pp,\; Qq$ be any two conjugate diameters, a parallelogram $DEGH$ formed by tangents to the conjugate hyperbolas at their vertices is equal to the rectangle contained by $Aa,\; Bb$, the two axes.

Let $Aa$, one of the axes, meet the tangent $PE$ in $K$; join $QK$, and draw $PL,\; QM$ perpendicular to $Aa$.

Because $CK\; :\; CA\; =\; CA\; :\; CL$ (11), and $CA\; :\; CB\; =\; CL\; :\; QM$ (1 Cor. 25), ex. aq. $CK\; :\; CB\; =\; CA\; :\; QM$, therefore $CK\; \backslash cdot\; QM\; =\; CB\; \backslash cdot\; CA$.

But $CK\; :\; QM\; =\; \backslash text\{twice\; trian.\; \}\; CPEQ$, therefore the parallelogram $CPEQ\; =\; CB\; \backslash cdot\; CA$; and, taking the quadruples of these, the parallelogram $DEGH$ is equal to the rectangle contained by $Aa$ and $Bb$.

If two tangents at the vertex of any transverse diameter of an hyperbola meet a third tangent; the rectangle contained by their segments between the points of contact and the points of intersection is equal to the square of the semidiameter to which they are parallel. And the rectangle contained by the segments of the third tangent between its points of contact and the parallel tangents, is equal to the square of the semidiameter to which it is parallel.

Let $PH,\; ph$, tangents at the vertices of a transverse diameter $Pp$, meet $DHh$, a tangent to the curve at any point $D$, in $H$ and $h$; let $CQ$ be the semidiameter to which

Hyperbola. the tangents $PH,\; ph$ are parallel, and $CR$ that to which $Hh$ is parallel; then

Let $Hh$ meet the semidiameters $CP,\; CQ$ in $L$ and $K$. Draw $DE,\; RM$ parallel to $CQ$, and $DG$ parallel to $CP$.

hence, and because of the parallels $PH,\; ED,\; CK,\; ph$,

wherefore $PH\; \backslash cdot\; ph\; =\; ED\; \backslash cdot\; CK$,

therefore $PH\; \backslash cdot\; ph\; =\; CQ^2$.

Again, the triangles $LED,\; CMR$ are evidently similar, and $LE,\; LD$ are similarly divided at $H$ and $P$, also at $h$ and $p$;

hence, taking the rectangles of the corresponding terms,

But if $CD$ be joined, the points $D$ and $R$ are evidently the vertices of two conjugate diameters (Cor. Def. 15), and therefore $PE\; \backslash cdot\; pE\; =\; CM^2$ (25);

COR. The rectangle contained by $LD$ and $DK$, the segments of a tangent intercepted between $D$ the point of contact, and $Pp,\; Qq$, any two conjugate diameters, is equal to the square of $CR$, the semidiameter to which the tangent is parallel.

Let the parallel tangents $PH,\; ph$ meet $LK$ in $H$ and $h$, and draw $DE$, a semi-ordinate to $Pp$. Because of the parallels $ED,\; PH,\; CK,\; ph$,

therefore

PROP. XXVIII.

If two straight lines be drawn from the foci of an hyperbola perpendicular to a tangent; straight lines drawn from the centre, to the points in which they meet the tangent, will each be equal to half the transverse axis.

Let $PdD$ be a tangent to the curve at $P$, and $FD,\; fd$ perpendiculars to the tangent from the foci; the straight lines joining the points $C,\; D$ and $C,\; d$ are each equal to $AC$, half the transverse axis.

Join $FB,\; fP$, and produce $FD,\; Pf$ till they intersect in $E$. The triangles $FDP,\; EDP$ have the angles at $D$ right angles, and the angles $FDP,\; EPD$ equal (5), and the side $DP$ common to both; they are therefore equal, and con-

sequently have $ED\; =\; DF$, and $EP\; =\; PF$; wherefore $EF\; =\; EP\; -\; Pf\; =\; Aa$. Now the straight lines $FE,\; Ff$ being bisected at $D$ and $C$, the line $DC$ is parallel to $Ef$, and thus the triangles $FfE,\; FCD$, are similar,

therefore $Ff\; :\; fE$, or $Aa\; =\; FC\; :\; CD$;

but $FC$ is half $Ef$, therefore $CD$ is half of $Aa$.

COR. If a straight line $Qq$ be drawn through the centre parallel to the tangent $Dd$, it will cut off from $PF,\; Pf$ the segments $PG,\; Pg$, each equal to $AC$ half the transverse axis. For $CdPG,\; CDPg$ are parallelograms, therefore $PG\; =\; dC\; =\; AC$, and $Pg\; =\; DC\; =\; AC$.

PROP. XXIX.

The rectangle contained by perpendiculars drawn from the foci of an hyperbola to a tangent is equal to the square of half the conjugate axis.

(See Figure to Prop. XXVIII.)

Let $PdD$ be a tangent, and $FD,\; fd$ perpendiculars from the foci, the rectangle contained by $FD$ and $fd$ is equal to the square of $BC$ half the conjugate axis.

It is evident from last proposition that the points $D,\; d$, are in the circumference of a circle, whose centre is the centre of the hyperbola, and radius $CA$ half the transverse axis; now $FDd$ being a right angle, if $dC$ be joined and produced, it will meet $DF$ in $H$, a point in the circumference; and since $FC\; =\; fC$, and $CH\; =\; Cd$, and the angles $FCH,\; fCd$ are equal, $FH$ is equal to $fd$, therefore $Df\; \backslash cdot\; df\; =\; DF\; \backslash cdot\; FH\; =\; AF\; \backslash cdot\; aF$ (36, 3, E.) $=\; CB^2$ (3).

COR. If $PF,\; Pf$ be drawn from the point of contact to the foci; the square of $FD$ is a fourth proportional to $fP,\; FP$, and $CB^2$. For the angles $fPd,\; FPD$ are equal (4), and $FDp,\; fdP$ are right angles, therefore the triangles $FDp,\; fdP$ are similar, and

PROP. XXX.

If from $C$ the centre of an hyperbola a straight line $CL$ be drawn perpendicular to a tangent $LD$, and from $D$ the point of contact a perpendicular be drawn to the tangent, meeting the transverse axis in $H$ and the conjugate axis in $h$; the rectangle contained by $CL$ and $DH$ is equal to the square of $CB$, the semiconjugate axis; and the rectangle contained by $CL$ and $Dh$ is equal to the square of $CA$, the semitransverse axis.

Let the axes meet the tangent in $M$ and $m$, and from $D$ draw the semi-ordinates $DE,\; Dc$, which will be perpendicular to the axes.

The triangles $DEH,\; CLm$, are evidently equiangular,

In the same way it may be shown that $CL\; \backslash cdot\; Dh\; =\; AC^2$. COR. 1. If a perpendicular be drawn to a tangent at the point of contact; the segments intercepted between the point of contact and the axes are to each other reciprocally as the semi-ordinates.

cally as the squares of the axes by which they are terminated.

For $AC^2\; :\; BC^2\; =\; CL\; \backslash cdot\; Dh\; :\; CL\; \backslash cdot\; DH\; =\; Dh\; :\; DH$.

Cor. 2. If $DF$ be drawn to either focus, and $HK$ be drawn perpendicular to $DF$; the straight line $DK$ shall be equal to half the parameter of the transverse axis.

Draw $CG$ parallel to the tangent at $D$, meeting $DH$ in $N$, and $DF$ in $G$. The triangles $GDN$, $HDK$, are similar, therefore $GD\; :\; DN\; =\; HD\; :\; DK$; and hence $GD\; \backslash cdot\; DK\; =\; HD\; \backslash cdot\; DN$.

But $GD\; =\; AC$ (Cor. 28) and $ND\; =\; CL$, therefore $AC\; \backslash cdot\; DK\; =\; HD\; \backslash cdot\; CL\; =$ (by the Prop.) $CB^2$, wherefore $AC\; :\; BC\; =\; BC\; :\; DK$; hence $DK$ is half the parameter of $Aa$ (Def. 16).

(See Figure to next Prop.)

XVII. If a point $G$ be taken in the transverse axis of an hyperbola, so that the distance of $G$ from the centre may be a third proportional to $CF$, the distance of either focus from the centre, and $CA$ the semitransverse axis; a straight line $HGa$ drawn through $G$, perpendicular to the axis, is called the directrix of the hyperbola.

Cor. 1. If an ordinate to the axis be drawn through the focus; tangents to the hyperbola at the extremities of the ordinate will meet the axis at the point $G$ (11).

Cor. 2. The hyperbola has two directrices, for the point $G$ may be taken on either side of the centre.

The distance of any point in an hyperbola from either directrix is to its distance from the focus nearest that directrix, in the constant ratio of the semitransverse axis to the distance of the focus from the centre.

Let $D$ be any point in the hyperbola; let $DK$ be drawn perpendicular to the directrix, and $DF$ to the focus nearest the directrix; $DK$ is to $DF$ as $CA$, half the transverse axis, to $CF$, the distance of the focus from the centre.

Draw $Df$ to the other focus, and $DE$ perpendicular to $Aa$; take $L$ a point in the axis so that $AL\; =\; FD$, and consequently $La\; =\; Df$; then $CL$ is evidently half the sum of $AL$ and $aL$, or of $FD$ and $fD$, and $CE$ half the sum of $FE$ and $fE$; and because

$Df\; -\; DF\; :\; fE\; =\; fE\; +\; FE\; :\; Df\; +\; DF$ (K, 6, E.), by taking the halves of the terms of the proportion,

But $CA\; :\; CF\; =\; CG\; :\; CA$ (Def. 17), therefore $CG\; :\; CA\; =\; CE\; :\; CL$; hence (19, 5, E.) $EG\; :\; AL\; =\; CG\; :\; CA\; =\; CA\; :\; CF$, that is, $DK\; :\; DF\; =\; CA\; :\; CF$.

Cor. 1. If the tangent $GMN$ be drawn through $M$, the

extremity of the ordinate passing through the focus, and Hyperbola. $ED$ be produced to meet $GM$ in $N$; $EN$ shall be equal to $DE$. For draw $MO$ perpendicular to the directrix, then, because $M$ and $D$ are points in the hyperbola, and from similar triangles,

Cor. 2. If $AI$ and $ai$ be drawn perpendicular to the transverse axis at its extremities, meeting the tangent $GM$ in $I$ and $i$, then $AI\; =\; AF$ and $ai\; =\; aF$.

If through $P$ and $Q$ the vertices of two semidiameters of an hyperbola there be drawn straight lines $PD$, $QE$ parallel to one of the asymptotes $CN$, meeting the other asymptote in $D$ and $E$; the hyperbolic sector $PCQ$ is equal to the hyperbolic trapezium $PDEQ$.

Let $CQ$ meet $PD$ in $I$. The triangles $CDP$, $CEQ$ are equal (1 Cor. 17); therefore, taking the triangle $CDI$ from both, the triangle $CIP$ is equal to the quadrilateral $DEQI$. To these add the figure $PIQ$, and the hyperbolic sector $PCQ$ is equal to the hyperbolic trapezium $PDEQ$.

If from the centre of an hyperbola the segments $CD$, $CE$, $CH$ be taken in continued proportion in one of the asymptotes, and the straight lines $DP$, $EQ$, $HR$ be drawn parallel to the other asymptote, meeting the hyperbola in $P$, $Q$, $R$; the hyperbolic areas $PDEQ$, $QEHR$ are equal.

(See Figure to preceding Prop.)

Through $Q$ draw a tangent to the curve, meeting the asymptotes in $K$ and $L$; join $PR$, meeting the asymptotes in $M$ and $N$; draw the semidiameters $CP$, $CQ$, $CR$; let $CQ$ meet $PR$ in $G$.

Because $QE$ is parallel to $CM$, and $KQ$ is equal to $QL$ (2 Cor. 16), $CE$ is equal to $EL$; and because $MC$, $PD$, $RH$, are parallel, and $MP$ is equal to $RN$ (1 Cor. 16), $CD$ is equal to $HN$. Now, by hypothesis,

and, by alternation, $NH\; :\; HR\; =\; LE\; :\; EQ$.

Now the angles at $H$ and $E$ are equal, therefore the triangles $NHR$, $LEQ$ are equiangular, and $NR$ is parallel to $LQ$; consequently $RP$ is an ordinate to the diameter $CQ$ (10), and is bisected by it at $G$; and as $CQ$ bisects all lines which are parallel to $KL$, and are terminated by the hyperbola, it will bisect the area $PQR$. Let the equal areas $PQG$, $RQG$ be taken from the equal triangles $PCG$, $RCG$, and there will remain the hyperbolic sectors $PCQ$, $RCQ$ equal to each other. Therefore (32) the areas $DPQE$, $EQRH$ are also equal.

Cor. Hence if $CD$, $CE$, $CH$, &c. any number of segments of the asymptote, be taken in continued proportion, the areas $DPQE$, $DPQRH$, &c. reckoned from the first line $DP$, will be in arithmetical progression.

Two straight lines $Aa$, $Bb$, which bisect each other at right angles in $C$, being given by position; to describe an hyperbola, of which $Aa$ shall be the transverse and $Bb$ the conjugate axes.

Join $AB$, and in $Aa$ produced take $CF$, $Cf$ each equal to $AB$; the points $F$, $f$ will be the foci of the hyperbola.

Let one end of a string be fastened at $F$, and the other to $G$ the extremity of a ruler $fDG$, and let the difference between the length of the ruler and the string be equal to $Aa$. Let the other end of the ruler be fixed to the point $f$, and let the ruler be made to revolve about $f$ as a centre in the plane in which the axes are situated, while the string is stretched by means of a pin $D$, so that the part of it between $G$ and $D$ is applied close to the edge of the ruler; the point of the pin will by its motion trace a curve line $DAD$ upon the plane, which is one of the hyperbolas required.

If the ruler be made to revolve about the other focus $F$, while the end of the string is fastened to $f$, the opposite hyperbola will be described by the moving point $D$; for in either case $Gf\; =\; (GD\; +\; DF)$, that is, $Df\; =\; DF$ by hypothesis equal to $Aa$ the transverse axis.

An hyperbola being given by position; to find its axes.

Let $HAH$ be the given hyperbola. Draw two parallel straight lines $HL$, $KA$, terminating in either of the opposite hyperbolas, and bisect them at $L$ and $M$; join $LM$, and produce it to meet the hyperbola in $P$; then $LP$ will be a transverse diameter (4 Cor. 10). Let $p$ be the point in which it meets the opposite hyperbola, bisect $Pp$ in $C$, the point $C$ is the centre (2). Take $D$ any point in the hyperbola, and on $C$ as a centre with the distance $CD$ describe a circle; if this circle lie wholly without the opposite hyperbolas, then $CD$ must be half the transverse axis (23); but if not, let the circle meet the hyperbola again in $d$; join $Dd$, and bisect it in $E$; join $CE$, meeting the opposite hyperbolas in $A$ and $a$, then $Aa$ will be the transverse axis (5 Cor. 10); for it is perpendicular to $Dd$ (3, 3, E.), which is an ordinate to $Aa$. The other axis will be found by drawing $Bb$ a straight line through the centre perpendicular to $Aa$, and taking $CB$ so that $CB$ may be a fourth proportional to the rectangle $AE\; \backslash cdot\; Ea$, and the squares of $DE$ and $CA$; thus $CB$ is half the conjugate axis (22).

I. If through the point $V$, without the plane of the circle $ADB$, a straight line $AVE$ be drawn, and produced indefinitely both ways, and if the point $V$ remain fixed while the straight line $AVE$ is moved round the whole circumference of the circle; two superficies will be generated by its motion, each of which is called a Conical Superficies, and these mentioned together are called Opposite Conical Superficies.

II. The solid contained by the conical superficies, and the circle $ADB$, are called a Cone.

III. The fixed point $V$ is called the Vertex of the cone.

IV. The circle $ADB$ is called the Base of the cone.

V. Any straight line drawn from the vertex to the circumference of the base is called a Side of the cone.

VI. A straight line $VC$ drawn through the vertex of the cone, and the centre of the base, is called the Axis of the cone.

VII. If the axis of the cone be perpendicular to the base, it is called a Right cone.

VIII. If the axis of the cone be not perpendicular to the base, it is called a Scalene cone.

If a cone be cut by a plane passing through the vertex; the section will be a triangle.

Let $ADBV$ be a cone of which $VC$ is the axis; let $AD$ be the common section of the base of the cone and the cutting plane; join $VA$, $VD$. When the generating line comes to the points $A$ and $D$, it is evident that it will coincide with the straight lines $VA$, $VD$; they are therefore in the surface of the cone, and they are in the plane which passes through the points $V$, $A$, $D$, therefore the triangle $VAD$ is the common section of the cone and the plane which passes through its vertex.

If a cone be cut by a parallel to its base; the section will be a circle, the centre of which is in the axis.

Let EFG be the section made by a plane parallel to the base of the cone, and VAB, VCD two sections of the cone made by any two planes passing through the axis VC; let EG, HF be the common sections of the plane EFG, and the triangles VAB, VCD. Because the planes EFG, ADB are parallel, HE, HF will be parallel to CA, CD, and

$AC\; :\; EH\; =\; (VC\; :\; VH\; =)\; CD\; :\; HF$; but $AC\; =\; CD$, therefore $EH\; =\; HF$. For the same reason $GH\; =\; HF$, therefore EFG is a circle of which H is the centre and EG the diameter.

If a scalene cone ADBV be cut through the axis by a plane perpendicular to the base, making the triangle VAB, and from any point H in the straight line AV a straight line HK be drawn in the plane of the triangle VAB, so that the angle VHK may be equal to the angle VBA, and the cone be cut by another plane passing through HK perpendicular to the plane of the triangle ABC, the common section HFKN of this plane and the cone will be a circle.

Take any point L in the straight line HK, and through L draw EG parallel to AB, and let EFGN be a section parallel to the base, passing through EG; then the two planes HFKN, EFGN being perpendicular to the plane VAB, their common section FLN is perpendicular to ELG, and since EFGN is a circle (by last Prop.), and EG its diameter, the square of FL is equal to the rectangle contained by EL and LG (35, 3, E.); but since the angle VHK is equal to VBA or VGE, the angles EHK, EGK are equal, therefore the points E, H, G, K, are in the circumference of a circle (21, 3, E.), and $HL\; \backslash cdot\; LK\; =\; EL\; \backslash cdot\; LG$ (35, 3, E.) $=\; FL^2$, therefore the section HFKN is a circle of which HLK is a diameter (35, 3, E.)

This section is called a Subcontrary Section.

If a cone be cut by a plane which does not pass through the vertex, and which is neither parallel to the base nor to the plane of a subcontrary section; the common sec-

tion of the plane and the surface of the cone will be an ellipse, a parabola, or an hyperbola, according as the plane passing through the vertex parallel to the cutting plane falls without the cone, touches it, or falls within it.

Let ADBV be any cone, and let ONP be the common section of a plane passing through its vertex and the plane of the base, which will fall without the base, will touch it, or will fall within it.

Let FKM be a section of the cone parallel to VPO; through C the centre of the base draw CN perpendicular to OP, meeting the circumference of the base in A and B; let a plane pass through V, A, and B, meeting the plane OVP in the line NV, the surface of the cone in VA, VB, and the plane of the section FKM in LK; then, because the planes OVP, MKF are parallel, KL will be parallel to VN, and will meet VB one side of the cone in K; it will meet VA the other side in H, fig. 1, within the cone; it will be parallel to VA in fig. 2, and it will meet VA, produced beyond the vertex, in H, fig. 3.

Let EFGM be a section of the cone parallel to the base, meeting the plane VAB in EG, and the plane FKM in FM, and let L be the intersection of EG and FM; then EG will be parallel to BN, and FM will be parallel to PO, and therefore will make the same angle with LK, wherever the lines FM, LK cut each other; and since BN is perpendicular to PO, EG is perpendicular to FM. Now the section EFGM is a circle of which EG is the diameter (2), therefore FM is bisected at L, and $FL^2\; =\; EL\; \backslash cdot\; LG$.

CASE 1. Let the line PNO be without the base of the cone. Through K and H draw KR and HQ parallel to AB. The triangles KLG, KHQ are similar, as also HLE, HKR; therefore $KL\; :\; LG\; =\; KH\; :\; HQ$, and $HL\; :\; LE\; =\; KH\; :\; KR$;

therefore $KL\; \backslash cdot\; HL\; :\; LG\; \backslash cdot\; LE$ or $LF^2\; =\; KH^2\; :\; HQ\; \backslash cdot\; KR$. Now the ratio of $KH^2$ to $HQ\; \backslash cdot\; KR$ is the same wherever the sections HFKN, EFGM intersect each other; therefore $KL\; \backslash cdot\; HL$ has a constant ratio to $LF^2$, consequently (1 Cor. 13, Part II.) the section HFKN is an ellipse, of which HK is a diameter and MF an ordinate.

CASE 2. Next, suppose the line ONP to touch the circumference of the base in A. Let DIS be the common section of the base and the plane FKM; the line DIS is evidently parallel to FLM, and perpendicular to AB, therefore $DI^2\; =\; AI\; \backslash cdot\; IB$,

hence $DI^2\; :\; FL^2\; =\; AI\; \backslash cdot\; IB\; :\; EL\; \backslash cdot\; LG$.

But since EG is parallel to AB, and IK parallel to AV, AI is equal to EL, and

Hence it appears (Cor. 9, Part I.), that the section DFKMS is a parabola, of which KLI is a diameter, and DIS, FLM ordinates to that diameter.

CASE 3. Lastly, let the line PNO fall within the base; draw VT through the vertex parallel to EG. The triangles HVT, HEL are similar, as also the triangles KVT, KGL, therefore

therefore $HT\; \backslash cdot\; KT\; =\; TV^2\; =\; HL\; \backslash cdot\; LK\; :\; LE\; \backslash cdot\; LG$ or $LF^2$. Hence it appears that $HL\; \backslash cdot\; LK$ has to $LF^2$ a constant ratio, therefore the section DFKMS is an hyperbola, of which KH is a transverse diameter, and FM an ordinate to that diameter (2 Cor. 19, Part III.).

SCHOLIUM. From the four preceding propositions it appears, that the only lines which can be formed by the common section of a plane and the surface of a cone, are these five: 1. A straight line, or rather two straight lines intersecting each other in the vertex of the cone, and forming with the straight line which joins the points in which they meet the base, a triangle; 2. A circle; 3. An ellipse; 4. A parabola; 5. An hyperbola. The first two of these, however, viz. the triangle and circle, may be referred to the hyperbola and the ellipse; for if the axes of an hyperbola be supposed to retain a constant ratio to each other, and, at the same time, to diminish continually, till at last the vertices coincide; the opposite hyperbolas will evidently become two straight lines intersecting each other in a point; and a circle may be considered as an ellipse, whose axes are equal, or whose foci coincide with the centre; so that the only three sections which require to be separately considered are the ellipse, the parabola, and the hyperbola.

SECT. II.—OF THE CURVATURE OF THE CONIC SECTIONS.

DEFINITIONS.

I. A circle is said to touch a conic section in any point

when the circle and conic section have a common tangent at that point.

II. If a circle touch a conic section in any point, so that no other circle touching it in the same point can pass between it and the conic section on either side of the point of contact, it is said to have the same curvature with the conic section in the point of contact, and it is called the Circle of Curvature.

LEMMA.

If straight lines be drawn touching a circle at the extremities of any chord, and from any point in the circumference straight lines be drawn parallel to the tangents, to terminate in the chord; these lines will be equal; and the square of each will be equal to the rectangle contained by the segments of the chord between each line and the point of contact of the tangent to which it is parallel.

Let PL be any chord in a circle, and FPG, FLK tangents at P and L; if from any point H in the circumference there be drawn HE, HD parallel to FG, FL respectively, meeting the chord in E and D; the lines HE, HD are equal; and the square of each is equal to the rectangle $LD\; \backslash cdot\; PE$.

The sides of the triangle HDE being parallel to those of the triangle FLP, viz. HD to FL and HE to FP, the triangles are equiangular (29, 1, E.); now the angles FPL, FLP of the latter are equal (32, 3, E.), therefore the corresponding angles HED, HDE of the former are equal, and $HD\; =\; HE$.

And because the angle DHL is equal to HLK (29, 1, E.), and this last is equal to EPH (32, 3, E.); also, because the angle HLD is equal to HPG (32, 3, E.), which again is equal to PHE (29, 1, E.); the triangles DHL, EPH are equiangular. Therefore

and (16, 6, E.) $LD\; \backslash cdot\; EP\; =\; DH\; \backslash cdot\; HE\; =\; HD^2\; =\; HE^2$. If the point h be in the arc of the opposite segment, and hd be drawn parallel to FL, and hE to PF, it will in like manner appear that

COR. 1. The points D and d being determined as directed in the proposition,

The triangles DLH, HLP have the angle at L common to both, and the angles DHL, HPL equal, because each is equal to the angle HLK (29, 1, E. and 32, 3, E.), therefore they are equiangular; hence

and $LP\; :\; LD\; (=\; LP^2\; :\; LH^2)\; =\; LH^2\; :\; LD^2$ (Cor. 19, 6, E.). In the same way it may be proved that

COR. 2. If E, the intersection of the chords $Hh$, $PL$, be between P and L, the points D, $d$ will be on opposite sides of the point P. For in this case the chord $LH$ will be greater than the chord $LP$, and the chord $Lh$ will be less; therefore $LH^2$ will be greater than $LP^2$, and $Lh^2$ less; consequently (from Cor. 1) $LD$ will be greater than $LP$, and $Ld$ less than $LP$.

COR. 3. If the chord $Hh$, which is parallel to the tangent $PG$, be supposed to approach continually towards that tangent; the points D, $d$, and E will continually approach to P, and may come nearer to it than any assignable distance.

If a circle be described touching a conic section, and cutting off from the diameter that passes through the point of contact a segment greater than the parameter of that diameter; a part of the circumference on each side of the point of contact will be wholly without the conic section; but if it cut off a segment less than the parameter, a part of the circumference on each side of the point of contact will be wholly within the conic section.

CASE 1. Let the section be a parabola $KNPnk$; let $Pp$ be any diameter, and $PG$ a tangent at $P$ its vertex. Let a circle $KHPkh$ touch the parabola and tangent at $P$, and cut off from the diameter a segment $PL$ either greater or less than its parameter (fig. 1, 2). An arc $HPh$ of the circle, extending to each side of the vertex $P$, will be wholly without the parabola or wholly within it.

First let $PL$ be greater than $LR$, a segment of the diameter equal to the parameter (fig. 1); draw $LQ$ touching the circle at $L$; let $Nn$, an ordinate to the diameter $Pp$, meet the circle in $H$ and $h$, and draw $HD$, $hd$ parallel to $LQ$, meeting the diameter in $D$ and $d$.

Because $NE^2\; =\; nE^2\; =\; PE\; \backslash cdot\; LR$ (Cor. to 11 of Part I.), and $HE^2\; =\; PE\; \backslash cdot\; LD$ (Lemma); also $hE^2\; =\; PE\; \backslash cdot\; Ld$ (Lemma); therefore $NE^2\; :\; HE^2\; =\; LR\; :\; LD$, and $nE^2\; :\; hE^2\; =\; LR\; :\; Ld$.

Now since $Pd$ may be less than any given line (Cor. 3 to Lemma), let it be less than $PR$; then $LD$ and $Ld$ will both be greater than $LR$, and consequently $HE$ will be greater than $NE$, and $hE$ greater than $nE$; therefore the arc of the circle $HPHh$ will be wholly without the parabola.

Next let $PL$, the segment of the diameter cut off by the circle, be less than $PR$ the parameter; as before, let $Nn$ an ordinate to the diameter $Pp$ meet the circle in $H$ and $h$, and let $HD$, $hd$ be parallel to the tangent $LQ$; then, as in the preceding case,

Suppose now $PD$ to be less than $PR$ (Cor. 3, Lemma), then $LD$ and $Ld$ will each be less than $LR$, and therefore $HE$ will be less than $NE$, and $hE$ less than $nE$; hence the arc $HPHh$ of the circle will be within the parabola.

CASE 2. Now let the curve be an ellipse or hyperbola (fig. 3 and 4).¹ First let the circle, which has a common tangent $PG$ with the curve at the vertex of the diameter, cut off from it a segment $PL$ greater than $LR$, a segment of the chord equal to its parameter. Let $Nn$, an ordinate to $Pp$, meet the circle in $H$ and $h$; draw $LQ$ touching the circle at $L$, and draw $HD$, $hd$ parallel to $LQ$. Take a point $V$ in $Pp$, such that

Now $PE\; \backslash cdot\; Ep\; :\; NE^2\; =\; Pp\; :\; LR\; =\; Ep\; :\; LV$ (15, Part II., and 21, Part III.);

¹ The reasoning in the case of the hyperbola is exactly like that for the ellipse; therefore, to avoid multiplying figures, those for the hyperbola are omitted.

Curvature, hence also $PE\; \backslash cdot\; Ep\; :\; NE^2\; =\; PE\; \backslash cdot\; Ep\; :\; PE\; \backslash cdot\; LV$ (1, 6, E.),
therefore $NE^2\; =\; nE^2\; =\; PE\; \backslash cdot\; LV$.

Now $HE^2\; =\; PE\; \backslash cdot\; LD$, and $hE^2\; =\; PE\; \backslash cdot\; Ld$,
therefore $NE^2\; :\; HE^2\; =\; LV\; :\; LD$,
and $nE^2\; :\; hE^2\; =\; LV\; :\; Ld$.

Let the chord $HA$ of the circle have such a position that $Pd$ is less than $PR$ (Cor. 3 to Lemma); then $pd$ and $pD$ will both be greater than $pR$, and consequently greater than $pV$, which is less than $pR$. In this position of the chord, and in every other nearer to the tangent, $NE$ will be less than $HE$, and $nE$ less than $hE$; therefore the arc $HPA$ of the circle will be entirely without the ellipse or hyperbola on each side of their common point $P$.

Lastly, suppose that the segment $LP$ of the diameter cut off by the circle is less than $LR$ its parameter; then the same construction being made as in the other case, we shall have $NE^2\; :\; HE^2\; =\; LV\; :\; LD$,
$nE^2\; :\; hE^2\; =\; LV\; :\; Ld$.

Now, when the ordinate $Nn$ approaches toward the tangent $PG$, the point $D$ will approach to $P$, and the point $V$ to $R$; therefore there will be a position of the ordinate in which $LD$ and $Ld$ will be both less than $LV$; and the same will be true for all positions nearer to the tangent. In these, $NE$ will be greater than $HE$, and $nE$ greater than $hE$; thus it appears that the arc $HPA$ will be within the curve, to a certain extent, on each side of the point $P$.

COR. 1. If a circle touch a conic section, and cut off from the diameter that passes through the point of contact a segment equal to its parameter, it will have the same curvature with the conic section in the point of contact.

For if a greater circle be described, it will cut off a segment greater than its parameter, therefore a part of its circumference on each side of the point of contact will be wholly without the conic section; and as it will also be without the former circle, it will not pass between that circle and the conic section at the point of contact. If, on the other hand, a less circle be described, it will cut off from the diameter a less segment than its parameter, therefore a part of the circumference on each side of the point of contact will fall within the conic section; and as it will be within the former circle, it will not pass between that circle and the conic section at the point of contact. Hence (Def. 2), the circle which cuts off a segment equal to the parameter is the circle of equal curvature.

PROP. II.

The circle of curvature at the vertex of the axis of a parabola, or at the vertex of the transverse axis of an ellipse or hyperbola, falls wholly within the conic sec-

tion; but the circle of curvature at the vertex of the conjugate axis of an ellipse falls wholly without the conic section.

Fig. 1.

Let $Pp$ be the axis of a parabola, and $PHLa$ the circle of curvature at its vertex, which therefore (1 Cor. 1) cuts off from the axis a segment $PL$ equal to the parameter of the axis; if a tangent were drawn to the parabola at its vertex, it would also be a tangent to the circle at that point, therefore the centre of the circle is in $Pp$. Let $NE$, an ordinate to the axis, meet the circle in $H$ and $h$. It may be shown, as in the preceding proposition, that $NE^2\; :\; HE^2\; =\; LP\; :\; LE$.

Now, in every position of the ordinate, $LP$ is greater than $LE$; therefore $NE^2$ is always greater than $HE^2$, and $nE^2$ is greater than $hE^2$; therefore the circle is wholly without the parabola.

Next, let $Pp$ be the transverse axis of an ellipse (fig. 2), or hyperbola (fig. 3), or the conjugate axis of an ellipse (fig. 4), and $PHLa$ the circle of curvature; then, as in the parabola, the centre of the circle will be in the axis. In each case draw $Nn$, an ordinate to the axis meeting the circle in $H$ and $h$, and take a point $V$ in $PL$, so that $pP\; :\; pE\; =\; LP\; :\; LV$;

then it will appear, as in the last Proposition, that $NE^2\; :\; HE^2\; =\; LV\; :\; LE$.

Fig. 2.

Now, when $Pp$ is the transverse axis of an ellipse (fig. 2), since $Pp$ is greater than $LP$ and $Pp\; :\; PL\; =\; PE\; :\; PV$, therefore $PE$ is greater than $PV$; and hence $LV$ is always greater than $LE$; therefore $NE^2$ is greater than $HE^2$, also $nE^2$ is greater than $hE^2$; hence the circle falls wholly within the ellipse.

Fig. 3.

Again, when $Pp$ is the transverse axis of an hyperbola (fig. 3), $pE$ is greater than $pP$, and therefore $LV$ is greater than $LP$, and consequently greater also than $LE$; hence $NE^2$ is greater than $HE^2$, and $nE^2$ greater than $hE^2$, and the circle is wholly within the hyperbola.

Lastly, when $Pp$ is the conjugate axis of an ellipse (fig. 4), since $pP$ is in this case less than $LP$, and $pP\; :\; LP\; =\; PE\; :\; PV$, therefore $PE$ is less than $PV$; hence $LV$ is less than $LE$, and consequently $NE^2$ is less than $HE^2$, and $nE^2$ less than $hE^2$; therefore the circle is wholly without the ellipse.

The circle of curvature at the vertex of any diameter of a conic section which is not an axis, meets the conic section again in one point only; and between that point and the vertex of the diameter the circle falls wholly within the conic section on the one side, and wholly without it on the other.

CASE I. Let the conic section be a parabola, of which $PL$ is a diameter, and let $PLK$ be the circle of curvature at its vertex; cutting off from the diameter a segment $PL$ equal to its parameter; draw $PG$ touching the circle and parabola at $P$, and $LQ$ touching the circle in $L$; also draw $PK$ parallel to $LQ$, meeting the circle in $K$, and $KT$ parallel to $PG$, meeting the axis in $T$. The lines $KP$, $KT$ will be equal, and

therefore $KT$ is a semi-ordinate to the diameter $PL$ (Cor. Prop. 11, Part I.), and $K$ is a point in the parabola. And since only one line $PK$ can be drawn parallel to the tangent $LQ$, only one such point $K$ can be found; therefore the circle of curvature cuts the parabola in one point, be-

sides the vertex of the diameter, and in no more. Be Curvature. tween $P$ and $T$ draw $NE$ any ordinate to the diameter $PL$, meeting the circle in $H$ and $h$, and draw $HD$, $hd$ parallel to the tangents $LQ$, $HD$, meeting the axis in $D$ and $d$; and because

Now $PL$ is less than $LD$; therefore $NE$ is less than $HE$, and the circular arc $PHK$ is without the parabola from the vertex to the intersection $K$. If the ordinate be more remote from the vertex than the position $KT$, then $D$ and $d$ will be both on the same side of the vertex, and therefore $PL$ will be greater than $LD$, also greater than $Ld$, and consequently $NE$ will be greater than $HE$, also $nE$ than $hE$; hence it follows that the arc $PLK$ falls wholly within the parabola.

CASE 2. Let the conic section be either an ellipse or a hyperbola, of which $Pp$ is a diameter, and $PLK$ the circle of equal curvature at its vertex, cutting off a segment $PL$ equal to its parameter. Draw $LQ$ touching the circle, and $pQ$ touching the curve; and because this line is parallel to the line $PG$, which touches the circle and ellipse at $P$, the lines $LQ$, $pQ$ make equal acute or obtuse angles (but in opposite directions) with $pL$; therefore they will meet at a point $Q$. Join $PQ$; and because $Q$ is without the circle, and $P$ is in the circumference, the line $PQ$, which cannot be a tangent, must cut the circle in one other point $K$, and in no more. Draw $KS$ and $KT$ parallel to $QL$ and $Qp$, meeting $Pp$ in $S$ and $T$.

Because of the parallels,

therefore (by alt.) $Pp\; \backslash cdot\; PL\; =\; pT\; \backslash cdot\; LS\; =\; pT\; \backslash cdot\; TP\; :\; LS\; \backslash cdot\; TP$.

Hence $KT$ is a semi-ordinate to the diameter $Pp$, and $K$ is a point in the ellipse or hyperbola (15 of Part II. and 24 of Part III.).

Draw $NE$, any ordinate to the diameter, so as to meet the equicurve circle in $H$ and $h$, and the line $PQ$ in $Y$. Draw $HDZ$, $hdz$ parallel to $LQ$, meeting $Pp$, $PQ$ in $D$, $Z$ and in $d$, $z$; also draw $YV$ parallel to $LQ$, meeting $Pp$ in $V$.

Because of the parallel lines,

therefore $NE^2\; :\; HE^2\; =\; VL\; :\; LD\; =\; YK\; :\; KZ$,
and $nE^2\; :\; hE^2\; =\; VL\; :\; Ld\; =\; YK\; :\; Kz$.

Now wherever the point H be taken in the arc PHK, it is manifest that YK will be less than KZ, therefore also NE is less than HE; thus the arc PHK falls wholly without the conic section. Again, since YK always exceeds Kz, therefore nE always exceeds hE; hence the arc PHK falls wholly within the section.

The chord of the circle of curvature which is drawn from the point of contact through the focus of a parabola is equal to that which is cut off from the diameter; and half the radius of the circle is a third proportional to the perpendicular from the focus upon the tangent, and the distance of the point of contact from the focus.

Let PL be the chord cut off from the diameter, and PFH the chord passing through F the focus; draw PM the diameter of the circle; join HL, HM, and draw FK perpendicular to the tangent at P. Because the lines PFH, PL make equal angles with the tangent at P (3, Part I.), the angles PHL, PLH are equal (32, 3, E.); hence PH = PL. Secondly, the triangles FKP, PHM being manifestly similar,

$FK\; :\; FP\; =\; PH$, or $4\; PF\; :\; PM$,
hence $FK\; :\; FP\; =\; FP\; :\; \backslash frac\{1\}\{2\}\; PM$, or $\backslash frac\{1\}\{2\}$ the radius.

COR. 1. Hence the radius is equal to $\backslash frac\{2FP^2\}\{FK\}$.

COR. 2. The radius is also equal to $\backslash frac\{2FK^3\}\{AF^2\}$, where AF is the distance of the focus from the vertex of the parabola; for $FP\; =\; \backslash frac\{FK^2\}\{AF\}$ (13, Part I.).

COR. 3. Hence also the radius is equal to $\backslash frac\{\backslash frac\{1\}\{2\}\; L\; \backslash cdot\; FP^3\}\{FK^3\}$, where L denotes the parameter of the axis; for $\backslash frac\{2FP^2\}\{FK\}\; =\; \backslash frac\{2AF\; \backslash cdot\; FP^3\}\{AF^2\; \backslash cdot\; FP\; \backslash cdot\; FK\}\; =\; \backslash frac\{\backslash frac\{1\}\{2\}\; L\; \backslash cdot\; FP^3\}\{FK^3\}$.

The radius of the circle of curvature at the vertex of any diameter of an ellipse or hyperbola is a third proportional to the perpendicular drawn from the centre upon the tangent and half the conjugate diameter; and the chord which is drawn from the point of contact through the focus is a third proportional to the transverse axis and conjugate diameter.

Let PL be the chord cut off from the diameter, and PFH the chord passing through F the focus; draw PM the diameter of the circle, and from the centre O draw OR perpendicular to PL, which will bisect PL in R; join HM, and draw the conjugate diameter QCq meeting PH in N and PM in S, then PS is equal to the perpendicular from the centre C upon the tangent. The triangles PSC, PRO are similar; therefore

$PS\; :\; PC\; =\; PR\; :\; PO$;
but $PC\; :\; CQ\; =\; CQ\; :\; PR$ (Def. of param.),
therefore $PS\; :\; CQ\; =\; CQ\; :\; PO$.

Secondly, the triangles PSN, PHM are similar,
therefore $PN\; :\; PS\; =\; PM\; :\; PH$.

But $PS\; :\; CQ\; =\; (CQ\; :\; PO)\; =\; Cq\; :\; PM$,
therefore $PN\; :\; CQ\; =\; Cq\; :\; PH$,
or, since $PN\; =\; AC$ (Cor. to 19, Part II. and to 28, Part III.) $Aa\; :\; Cq\; =\; Cq\; :\; PH$.

COR. 1. Hence the radius of curvature is equal to $\backslash frac\{CQ^2\}\{PS\}$; and the chord passing through the focus is equal to $\backslash frac\{2CQ^2\}\{AC\}$.

COR. 2. The radius of curvature is also equal to $\backslash frac\{CQ^2\}\{AC\; \backslash cdot\; BC\}$; for $PS\; =\; \backslash frac\{AC\; \backslash cdot\; BC\}\{CQ\}$ (17, Part II. and 26, Part III.).

COR. 3. Draw FK from the focus perpendicular to the tangent, and let L denote the parameter of the transverse axis; the radius of curvature is also equal to $\backslash frac\{\backslash frac\{1\}\{2\}\; L\; \backslash times\; FP^3\}\{FK^3\}$.

For the triangles PEK, NPS are manifestly similar; therefore $FK\; :\; FP\; =\; PS\; :\; PN$, or $AC\; =\; BC\; :\; CQ$;

hence $CQ\; =\; \backslash frac\{FP\}\{FK\}\; \backslash times\; BC$, and

$\backslash frac\{CQ^3\}\{AC\; \backslash cdot\; BC\}\; =\; \backslash frac\{FP^3\}\{FK^3\}\; \backslash times\; \backslash frac\{BC^2\}\{AC\}\; =\; \backslash frac\{FP^3\}\{FK^3\}\; \backslash times\; \backslash frac\{1\}\{2\}\; L$.

This expression for the radius of curvature is the same for all the three conic sections.

Let ABCD, abcd, two trapeziums, have each two parallel sides, and let the angles which the parallel sides AB, DC of the one figure make with its side BC be equal to the angles which ab, dc, the parallel sides of the other figure, make with its side bc; also, in the one figure, let

EF, which is parallel to AB and DC, bisect the opposite sides BC, AD in F and E, and in the other figure let $ef$, a parallel to $ab$ and $dc$, bisect $bc$ and $ad$ in $e$ and $f$; the trapezium ABCD is to the trapezium $abcd$ as the rectangle BC·EF to the rectangle $bc·ef$.

Through E and e draw GH and $gh$ parallel to BC and $bc$, forming the parallelograms GC, $gc$. The triangles AEG, DEH are manifestly equal (26, 1, E.); therefore the trapezium ABCD is equal to the parallelogram GBCH. In the same way it appears that the trapezium $abcd$ is equal to the parallelogram $gbeh$. Now the parallelogram GC has to the parallelogram $gc$ the ratio compounded of the ratios of BG to $bg$, and of BC to $be$ (23, 6, E.); and the rectangle BC·BG has to the rectangle $bc·bg$ the ratio which is compounded of the same ratios; therefore $\backslash text\{trap.\; ABCD\}\; :\; \backslash text\{trap.\; \}\; abcd\; =\; BC\; \backslash cdot\; BG\; :\; bc\; \backslash cdot\; bg\; =\; BC\; \backslash cdot\; EF\; :\; bc\; \backslash cdot\; ef$.

In a parabola, let ABCD be a trapezium formed by PB, any diameter, AB, CD semi-ordinates to that diameter, and AD a chord in the curve; and let EFGH be another trapezium formed by PF, a tangent at the vertex of the diameter PB, by AF, DG, diameters produced at A, D, and EH a tangent parallel to the chord: the trapezium ABCD is double the trapezium EFGH.

Let L be the point of contact of the tangent EH; draw a diameter through L, meeting the chord AD in I, and the tangent PF in N; draw LM a semi-ordinate to the diameter PB, and IK parallel to LM. Let $p$ be the parameter of the diameter PB. And because AD is bisected in I, and $BK\; :\; KC\; =\; AI\; :\; ID\; =\; FN\; :\; NG$, therefore BC is bisected in K and FG in N.

And because $p\; \backslash cdot\; PB\; =\; AB^2\; =\; PF^2$,
and $p\; \backslash cdot\; PC\; =\; DC^2\; =\; PG^2$;
therefore $p\; \backslash cdot\; BC\; =\; PF^2\; -\; PG^2$
$=\; (PF\; +\; PG)(PF\; -\; PG)$,

that is, $p\; \backslash cdot\; BC\; =\; 2PN\; \backslash cdot\; FG\; =\; 2LM\; \backslash cdot\; FG$.

Now $p\; \backslash cdot\; PM\; =\; p\; \backslash cdot\; LN\; =\; LM^2$,
therefore $p\; \backslash cdot\; BC\; :\; p\; \backslash cdot\; LN\; =\; 2LM\; \backslash cdot\; FG\; :\; LM^2$,
and $BC\; :\; LN\; =\; 2FG\; :\; LM$ or $IK$,
and hence $BC\; \backslash cdot\; IK\; =\; 2FG\; \backslash cdot\; LN$.

Now, by the premised lemma, the trapezium ABCD is to the trapezium EFGH as the rectangle BC · IK to the rectangle FG · LN; therefore the trapezium ABCD is double the trapezium EFGH.

SCHOLIUM. Since GH may be of any magnitude, the proposition will still be true when the points H and G coincide in the line PF.

Let AB be a semi-ordinate to PB, any diameter of a parabola; complete the parallelogram ABPC, by drawing PC a tangent at the vertex, and AC parallel to PB: the space comprehended by PA, the arc of the parabola, and PB, AB, the absciss and ordinate, is two thirds of the parallelogram PBAC.

Divide PC into any odd number of equal parts, seven for example; let $Pd$ be one, $Pe$ three, and $Pf$ five of these; take $Pa\text{'}$ equal to $Pd$, and draw $dD\text{'}$, $eH$, $fK$ parallel to PB, meeting the curve in $D\text{'}$, $D$, $H$, $K$; these lines, when produced, will be diameters; and they will be equidistant, because they divide the line $Cd\text{'}$ into equal parts. Because $Pd\; =\; Pa\text{'}$, therefore $dD\; =\; dD\text{'}$ (11, 1, of this). Join $DD\text{'}$, which will be parallel to $da\text{'}$ (33, 1, E.); let it meet PB in N; produce it to meet the other diameters in E, F, G. Draw $HL$, $KM$ parallel to PC, also the chords $DH$, $HK$, $KA$, and, parallel to them, the tangents $dh$, $hk$, $ka$. And because the diameters whose vertices are $D\text{'}$, $D$, $H$, are equidistant, the tangents $da\text{'}$, $hd$ will intersect each other at $d$, a point in the diameter passing through $D$ (3 Cor. 15, 1); for a like reason the tangents $hd$, $hk$ will intersect in $h$, and $hk$, $ak$ in $k$.

The triangles $HED$, $hed$ are in all respects equal, for $DH\; =\; dh$ (34, 1, E.) $DE\; =\; de$; and since $Hh\; =\; Dd\; =\; Ec$, therefore $HE\; =\; he$. The trapeziums $KHEF$, $kh\; ef$ are also equal; for $HE$, a side of the one, is equal to $he$, a side of the other, and $HK$, $EF$ are equal to $hk$, $ef$ respectively, and make equal angles with the equal sides $HE$, $he$ between them; therefore if the trapeziums be applied, one on the other, so that the equal sides $HE$, $he$ coincide, the

sides $HK,\; hk$ will coincide, also $EF,\; ef$ and the trapeziums will entirely coincide. In like manner, the trapeziums $AKFG,\; akfc$ are proved to be equal; and because the trapezium $LHDN$ is double the triangle $hde$ or its equal $HDE$ (preceding Prop.), and the trapezium $MKHL$ double the trapezium $fkhe$, or its equal $FKHE$, and the trapezium $BAKM$ double $aCfk$, that is, double $AGFK$, the polygon $NDHKAB$ will be double the polygon $DHKAG$; but these together make up the parallelogram $ABNG$; therefore the polygon $NDHKAB$, inscribed in the parabola, is two thirds of the parallelogram $ABNG$. Now the space bounded by the arc $PA$, the absciss $PB$, and semiordinate $AB$, exceeds the inscribed polygon; therefore it also is greater than two thirds of the parallelogram $ABNG$, or two thirds of the parallelogram $ABPC$, diminished by two thirds of the parallelogram $PCGN$.

The parallelograms $HDdk,\; EDde$ are equal (35, 1, E.), so also are $KHhk,\; Eefc$, and $AKka,\; GF/c$ (36, 1, E.). Therefore the sum of the four parallelograms $Ak,\; Kh,\; Hd,\; DP$ is equal to the parallelogram $GNPC$. Now if the polygon inscribed in the parabola be increased by these four parallelograms, there will be formed the polygon $PdhkaB$, which exceeds the parabolic area; therefore that area is less than two thirds of the parallelogram $ABNG$ increased by the whole parallelogram $GNPC$, and consequently less than two thirds of the parallelogram $ABPC$ increased by one third of the parallelogram $GNPC$.

Let the parabolic space $PAB$ be denoted by $S$, the parallelogram $PCAB$ by $R$, and the parallelogram $PNGC$ by $V$.

It has been proved that $S\; >\; \backslash frac\{2\}\{3\}R\; -\; \backslash frac\{1\}\{3\}V$,
and that .

This is true whatever may be the magnitude of $V$; but the line $Pd$ may be taken such that $V$ may be less than any assignable space; therefore $S$ can be equal to no assignable space that is either greater or less than two thirds of the space $R$, and consequently is exactly equal to two thirds of the space $R$.

If the axes of a hyperbola be equal, it is called an Equilateral hyperbola.

If two ellipses, or two hyperbolas, have a common transverse axis, and if from the same point in the axis there be drawn a semiordinate to each; the areas contained by the common absciss, the ordinates, and the curves, will be to each other as their conjugate axes.

Let $AMBa,\; Amba$ be two ellipses, and $AMD,\; Amde$ two hyperbolas, which have each pair a common transverse axis $Aa$; and let $BC,\; bc$ be their conjugate axes, and $DE,\; de$ semiordinates at the same point $E$ in the common axis; the area $AMDE$ is to the area $Amde$ as the axis $BC$ to the axis $bc$.

Let the common absciss in both curves be divided into any number of equal parts $AP,\; PQ,\; QE$; draw semiordinates $PmM,\; QnN$, and the chords $AM,\; MN,\; ND,\; Am,\; mn,\; nd$.

In the case of two ellipses,

In the same way, in the two hyperbolas it may be proved that

Now, in both curves (by 1, 6, E.),
triangle $APM$ : triangle $APm$ = $PM\; :\; Pm$ = $BC\; :\; bC$;
and since $MP\; :\; NQ\; =\; mP\; :\; nQ$,
therefore, by composition and alternation,
$MP\; +\; NQ\; :\; mP\; +\; nQ\; =\; NQ\; :\; nQ\; =\; BC\; :\; bC$.
But $MP\; +\; NQ\; :\; mP\; +\; nQ\; =\; \backslash text\{trap.\; \}\; MPQN\; :\; \backslash text\{trap.\; \}\; mPQn$;
therefore, trap. $MPQN$ : trap. $mPQn$ = $BC\; :\; bC$.

In the same way it appears that the trapezium $NQED$ is to the trapezium $nQed$ as $BC$ to $bc$; therefore (12, 5, E.), polygon $AMNDE$ : polygon $Amnde$ = $BC\; :\; bC$.

This must be true, however great may be the number of sides of the polygon $AMNDE$, $Amnde$ inscribed in the curvilinear spaces; but by a known principle in geometry, the limits of the polygons (which are the curvilinear spaces) must have the same ratio as the polygons themselves; therefore the curvilinear spaces $AND,\; Amnde$ have the same ratio as the semiconjugate axes $BC,\; bC$.

COR. 1. Hence it appears that the area of an ellipse is to that of its circumscribing circle as the conjugate axis to the transverse axis.

COR. 2. It also appears, that the area of any segment of an ellipse may be found from that of a corresponding segment of a circle; and the area of a segment of any hyperbola from the corresponding segment of an equilateral hyperbola.

The quadrature of the hyperbola was perfectly accomplished by Archimedes: but it has been found impossible to find rectilinear spaces exactly equal to elliptic and hyperbolic areas; these can only be exhibited numerically by infinite series. In what follows, series will be investigated from principles as elementary as the nature of the subject will admit of; but it will be convenient to employ the symbols of algebra in the reasoning; and we shall confine the investigation to sectors of a circle and an equilateral hyperbola, from which sectors of any ellipse and hyperbola may be found.

Let $C$ be the centre of a circle, or of an equilateral hyperbola, and $CA$ the semitransverse axis; let $ACB,\; BCD$ be two equal sectors of the circle or hyperbola, and let $AHK$, a tangent at the vertex $A$, meet the semidiameters $CB,\; CD$ in $H$ and $K$.

In the circle $2CA\; :\; AK\; =\; CA^2\; -\; AH^2\; :\; CA\; \backslash cdot\; AH$.
In the hyperbola $2CA\; :\; AK\; =\; CA^2\; +\; AH^2\; :\; CA\; \backslash cdot\; AH$.

In each curve draw $DA,\; Da$ to the extremities of the axis; let the chord $AD$ meet the semidiameter $CB$ in $O$, and draw $DE$ perpendicular to the axis.

Because $AO\; =\; OD$ (3, 3, E. and Prop. 33, Part III.), and $AC\; =\; Ca$; the lines $CH,\; aD$ are parallel (2, 6, E.), therefore the triangles $CAH,\; aED$ are similar.

Now $ED^2\; =\; AE\; \backslash cdot\; Ea$ (35, 3, E. and 22 of Sect. III.), therefore $AE\; :\; ED\; =\; ED\; :\; Ea$ (16, 6, E.); hence the triangles $DEA,\; aED$ are similar (6, 6, E.), and each is similar to $CAH$, therefore

Hence, in the circle,

But $2CE\; :\; ED\; =\; 2CA\; :\; AK$; therefore, in the circle, $2CA\; :\; AK\; =\; AC^2\; -\; AH^2\; :\; AC\; \backslash cdot\; AH$; and, in the hyperbola, $2CA\; :\; AK\; =\; AC^2\; +\; AH^2\; :\; AC\; \backslash cdot\; AH$.

To investigate a formula that shall express the area of any sector of a circle, or of an equilateral hyperbola.

Let $ACD$ be a sector of a circle or hyperbola, and $AC$ the radius of the circle, or semitransverse axis of the hyperbola; let the sector be bisected by the semidiameter $CB$, and again each of the sectors $ACB,\; BCD$ by the semidiameters $CP,\; CQ$, and so on; thus dividing the sector $ACB$ first into two equal sectors, then into four, then into eight, and so on. Draw tangents to the curves at the alternate points $A,\; B,\; D$, and because, from the nature of the curves, the chords which join these points are ordinates to the semidiameters $CP,\; CQ$ which pass between them, the tangents will intersect each other at $G$ and $L$, points in the semidiameters (1 Cor. Prop. 10, Sect. III.). Join $A,\; B$, any two contiguous points, and let the chord $AB$ meet $CP$ in $I$; and because the triangles $ACI,\; BCI$ are equal (38, 1, E.), and also the triangles $AGI,\; BGI$, the triangles $CAG,\; CBG$ are equal. In the same way it appears that the triangles $DCL,\; BCL$ are equal.

Again, because the semidiameter $CB$ bisects the chord which joins the points $P,\; Q$, it will bisect $GL$, the tangent parallel to the chord; therefore the triangles $GCB,\; LCB$ are equal. Hence it appears that the triangles $ACG,\; GCB,\; BCL,\; LCD$ are all equal; and this will be true whatever be their number.

Let $AK$ a tangent at the vertex $A$, meet the semidiameters $CP,\; CB,\; CD$ in $G,\; H,\; K$. Put $a$ to denote $CA$, the radius of the circle or semiaxis of the hyperbola, and let the lines be denoted by These lines in the circle are the tangents of the arcs We shall, by analogy, consider them as tangents corresponding to the sectors $DCA,\; BCA,\; PCA$ in the two curves.

And because

the upper sign applying to the circles, and the lower to the hyperbola, therefore

Let the value of $\backslash frac\{a^2\}\{2t\text{'}\}$, found from this second formula be substituted instead of it in the first, and there is obtained

Now $4at\text{'}$ or $4CA\; \backslash cdot\; AG$ is eight times the triangle $ACG$, that is, double the polygon $CAGLD$. Let $s\text{'}$ denote this polygon, and then we have

By carrying on the process of bisecting the sector $ACD$ continually, so as to divide it next into 8, then into 16, then into 32 equal parts, and so on, putting $t\text{'}\text{'}$ for the tan-

gent corresponding to its 8th part; $t^8$ for that corresponding to its 16th, &c. there is obtained

Now the polygon CAGLD manifestly approaches continually to the area of the sector in either curve, as the number of its sides is increased, and at last may differ from it by less any than assignable quantity; therefore, putting now $s$ for the area of the sector ACB, and considering that $s$ is the limit of the polygon, we have

the series in the parenthesis being carried on ad infinitum; and hence, putting

This is the formula which was to be investigated, and in its application the upper part of the sign $\backslash pm$ applies to the circle, and the lower to the hyperbola.

SCHOLIUM. In computing the area of a sector of a circle, or hyperbola, the values of the tangents AH, AG, &c. that is, must be found from the first $t$, and from each other. In the circle

From this, by the resolution of a quadratic equation, we find

A few of the quantities may be computed from this formula; but when one ($t\text{'}$, for instance) has been found such that $\backslash frac\{t\text{'}\}\{a\}$ is a small fraction, those which follow may be most readily computed by a series obtained from $\backslash sqrt\{\backslash left(1\; +\; \backslash frac\{t\text{'}\}\{a\}\backslash right)\}$ by evolution, or the binomial theorem; thus, since

For the quadrature of the hyperbola, the tangents are found from the first by formulas entirely similar to the above, differing only in the signs of the terms. In this curve, however, the tangents have a property which those in the circle have not, by which their computation may be facilitated. For since, in the hyperbola,

This series of fractions, by which the tangents are expressed, being formed in a very simple manner from the square, the fourth, the eighth root, &c. of $v\; =\; \backslash frac\{a+t\}\{a-t\}$ they may be easily computed, and thence the values of , and the sector, readily found.

The trigonometrical tables may be made available in the quadrature of the hyperbolic sector, by the preceding formula in this way.

It has been proved in the calculus of sines, that $v$ and $z$ being any two arcs,

Suppose now that $v$ is a right angle $=\; \backslash frac\{1\}\{2\}\backslash pi$, then the formula becomes

Now we have, in the hyperbola,

From this it appears that $\backslash frac\{t\}\{a\}$ is related to $\backslash frac\{t\text{'}\}\{a\}$ exactly as in the trigonometrical formula $\backslash tan.\; \backslash frac\{1\}\{2\}z$ is related to $\backslash sin.\; z$;

and a like remark may be made on the relations between $\backslash frac\{t\text{'}\}\{a\}$ & $\backslash frac\{t\text{'}\text{'}\}\{a\}$.

Hence we have the following formula for finding the area of the hyperbolic sector.

Find from the trigonometrical tables the series of sines such that $\backslash sin.\; z\; =\; \backslash frac\{t\}\{a\}$

; then $s$ denoting, as above, the hyperbolic sector,

Since the series in the denominator approaches continually to a geometrical series, of which the common ratio is $\backslash frac\{1\}\{2\}$, any term is nearly three times the sum of all the terms which follow it. If then a term be found which agrees in all the figures with $\backslash frac\{1\}\{2\}$ of that before it, then $\backslash frac\{1\}\{2\}$ of that term may be taken as the sum of the remaining terms.

These expressions, and others, for the areas of the circle and hyperbola, were given by Professor Wallace of Edinburgh, in the Edinburgh Philosophical Transactions, vol. v. They are also contained in his Treatise on Conic Sections. (n.)