CONIC SECTIONS

IN treating of so considerable a branch of the mathematical sciences as the Conic Sections, it would be improper to pass over in total silence the history of those remarkable curves. But this topic will not require any long detail. None of the works of the more early Greek geometers have reached our time; nor have we any work of antiquity professedly written on the subject of our inquiry. Our curiosity must therefore rest satisfied with the knowledge of a few incidental 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 preferred by Euphr. * In Archim. *, and are attributed by him to Menechmus, 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 †, has been thought sufficient authority, by some authors, to ascribe the honour of the discovery of the conic sections to Menechmus. 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

same, in the school of Alexandria, under the successors of Eudoxus. 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.

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 farther 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 first 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 Pergaei 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 Ezechielensis, 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, was yet found to contain only the first seven books: and it is remarkable, that another manuscript, brought from the east by Gottus, the learned professor of Leyden, so early as 1664, as well as a third, of which Roxius 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 Collections Mathematicae of Pappus, in whose time the entire treatise of Apollonius was extant, there is preserved.

* In Archim. *, and are attributed by him to Menechmus, 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 †, has been thought sufficient authority, by some authors, to ascribe the honour of the discovery of the conic sections to Menechmus. We may at least infer that, at this epoch, geometers had made some progress in developing the properties of these curves.

† 182.

served 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 four last 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 connection 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 the 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 1655, 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 matter to conceive steadily, and to understand diagrams rendered confused by lines drawn in different planes: all which 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 lately published by 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 1st book of Euclid's Elements: (2 cor. 20. 6. E.) means the 2d corollary to the 20th prop. of the 6th 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 2d 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.

PART I. OF THE PARABOLA.

Plate
CLVI.

DEFINITIONS.

Fig. 1.

I. Is a straight line BC, and a point without it F, be given by 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.

COROLLARY. The lines DF, DB, may become greater than any given line; therefore the parabola

extends to a greater distance from the point F, and the line BC, than any that can be assigned.

II. The straight line BC, which is given by 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.

COR. 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 Abscissa.

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 DA 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.

COROLLARY. 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.

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 DE, 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. 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.

COR. 2. 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, and therefore is less than RF, the distance of the same point from the focus.

COR. 3. 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 shewn that FB, which is perpendicular to the tangent DI, is bisected at I.

COR. 4. 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. 5. 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.

From the property of tangents to the parabola demonstrated in Cor. 4. 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.

A straight line drawn from the focus of a parabola to the intersection of two tangents to the curve, will make equal angles with straight lines drawn from the focus to the points of contact.

LET $HP,\; Hp$ be tangents to a parabola at the points $P,\; p$; let a straight line be drawn from $H$, their intersection, to $F$ the focus, and let $FP,\; Fp$ be drawn to the points of contact, the lines $PF$ and $pF$ make equal angles with $HF$.

Draw $PK,\; pK$ perpendicular to the directrix; join $HK,\; Hk$, join also $FK,\; Fk$, meeting the tangents in $G$ and $g$. The triangles $FPH,\; KPH$ have $PF$ equal to $PK$, and $PH$ common to both, also the angle $FPH$ equal to $KPH$ (3.), therefore $FH$ is equal to $KH$, and the angle $HFP$ is equal to the angle $HKP$. In like manner it may be shewn that $FH$ is equal to $KH$, and that the angle $HFP$ is equal to the angle $Hkp$; therefore $HK$ is equal to $Hk$, and hence the angle $HKF$ is equal to $HkF$: now the angles $PKF,\; pKF$ are right angles, therefore the angle $HKP$ is equal to $Hkp$; but these angles have been shewn to be equal to $HFP$ and $Hkp$ respectively, therefore the lines $PF$ and $pF$ make equal angles with $HF$.

Two tangents to a parabola, which are limited by their mutual intersection and the points in which they touch the curve, are to each other reciprocally as the sines of the angles they contain with straight lines drawn from the points of contact to the focus.

LET $HP,\; Hp$, which intersect each other at $H$, be tangents to a parabola at the points $P,\; p$; and let $PF,\; pF$ be drawn to the focus: then

Join $HF$; and in $FP$ take $FQ$ equal to $Fp$, and join $HQ$; then, the angles at $F$ being equal (4.), the triangles $HFQ,\; HFp$ are equal, therefore $HQ$ is equal to $Hp$, and the angle $HQF$ is equal to $HpF$. Now, in the triangle $HPQ$,

Any straight line terminated both ways by a parabola, and parallel to a tangent, is bisected by the diameter that passes through the point of contact; or is an ordinate to that diameter.

The straight line $Dd$, terminated by the parabola, and parallel to the tangent $HPb$, is bisected at $E$ by $PE$ the diameter that passes through $P$ the point of contact.

Let $KD,\; Kd$, tangents at the points $D,\; d$, meet the tangent at the vertex in $H$ and $h$; draw $DL,\; dl$, parallel to $EP$, meeting $Hb$ in $L$ and $l$, and draw $DF,\; dF$, $PF$ to the focus.

Because $Hb$ is parallel to $Dd$,

But $KD,\; Kd$ being tangents to the parabola,

Therefore, $\backslash text\{sine\; \}\; bdf\; :\; \backslash text\{sine\; \}\; HDF\; ::\; HD\; :\; bd$;

Therefore, (23. 5. E.) $\backslash text\{sine\; \}\; bPF\; :\; \backslash text\{sine\; \}\; HDF\; ::\; HD\; :\; bP$; but $\backslash text\{sine\; \}\; HPF$, or $\backslash text\{sine\; \}\; bPF\; :\; \backslash text\{sine\; \}\; HDF\; ::\; HD\; :\; HP$, therefore the ratio of $HD$ to $bP$ is the same as that of $HD$ to $HP$, wherefore $HP\; =\; Pb$.

Again, because the angles $HDF$ and $bdf$ are respectively equal to $HDL$ and $hdl$, (3.)

Now $HL\; :\; DH\; ::\; \backslash text\{sine\; \}\; HDL\; :\; \backslash text\{sine\; \}\; HLD$, or $\backslash text\{sine\; \}\; hld$ (by Trigon.)

therefore (23. 5. E.) $HL\; :\; db\; ::\; \backslash text\{sine\; \}\; bdl\; :\; \backslash text\{sine\; \}\; hld\; ::\; bl\; :\; db$, wherefore $HL$, and $bl$, have the same ratio to $db$, hence $HL\; =\; bl$; and since it has been shewn that $HP\; =\; Pb$, it follows that $LP\; =\; Pl$, and therefore $DE\; =\; Ed$.

COR. 1. Straight lines which touch a parabola at the extremities of an ordinate to a diameter intersect each other in that diameter. For $Dd$ and $Hb$ being similarly divided at $E$ and $P$, the straight line which joins the points $E,\; P$, will pass through $K$ the vertex of the triangle $DKd$.

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 the axis are perpendicular to it, and no other diameter is perpendicular to its ordinates. This is evident from 2 cor. and 5 cor. to Prop. III.

COR. 6. Hence the axis divides the parabola into two parts which are similar to each other.

If a tangent to 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 the tangent is equal to the segment between the vertex and the ordinate.

LET $DK$, a tangent to the curve at $D$, meet the diameter $EP$ in $K$, and let $DEd$ be an ordinate to that diameter, $PK$ is equal to $PE$.

Through $P$, the vertex of the diameter, draw the tangent $PH$, meeting $KD$ in $H$; draw $DL$ parallel to $EP$, meeting $PH$ in $L$, and draw $DF,\; PF$ to the focus: then

PH : HD :: fine HDF : fine HPF (5.)
But the angle HDF is equal to HDL, and HPF is
equal to HPK (3.), that is (because of the parallel
lines DL, PK) to HLD, therefore

PH : HD :: fine HDL : fine HLD :: HL : HD,
wherefore PH = HL, and consequently PK = DL;
but PL is parallel to DE, by last proposition, and
therefore DL = PE, therefore PK = PE.

If an ordinate to any diameter passes 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 fo-
cus, be an ordinate to the diameter PE; the absciss
PE is equal to $\backslash frac\{1\}\{4\}$ the parameter, and the ordinate Dd
is equal to the whole parameter of the diameter PE.

Let DK, PI be tangents at D and P; let DK
meet the diameter in K; draw PF to the focus, and
DL parallel to EP. The angles KPI, IPF being
equal (3.), and PI parallel to EF (2 cor. 6.), the
angles PEF, PFE are also equal (29. i. E.), and
PE = PF = $\backslash frac\{1\}\{2\}$ the parameter (Def. 9. and Def. 1.).
Again, the angle KDE is equal to LDK (3.), and
therefore equal to DKE; consequently ED is equal
to EK, or to twice EP (7.); 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

Let PN, a semi-ordinate to the diameter HQ, meet
KR in O, and let PR, a semi-ordinate to the dia-
meter KO, meet HN in R; 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.

Therefore, by subtraction, $LI\; =\; MR\; =\; HQ$,

Therefore by addition, $IO\; =\; GQ$.

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

Therefore $GN\; :\; IO$, or $2GH\; :\; IO\; ::\; PG\; :\; PI$,
And $GQ\; :\; IR$, or $IO\; :\; 2IK\; ::\; PG\; :\; PI$,
Hence, taking the rectangles of the corresponding
terms,

Cor. The squares of semi-ordinates, and of ordinates
to any diameter, are to one another as their correspond-
ing abscisses. Let HEb, KNc be ordinates to the dia-

meter PN; draw PG a tangent to the curve at the
vertex of the diameter, and complete the parallelo-
grams PEHG, PNKI; then PG, PI are equal to
EH, NK, and GH, IK to PE, PN, respectively;
therefore HE' : KN' :: PE : PN.

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

Let HBb be an ordinate to the diameter PK, the Fig. 1.
rectangle contained by PB and the parameter of the
diameter is equal to the square of HB, 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 (8.).

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

It was on account of the equality of the square of
the semi-ordinate to a rectangle contained by the pa-
rameter of the diameter and the absciss, that Apol-
lonius called the curve line to which the property be-
longed a Parabola.

A straight line drawn from the focus of a para-
bola, 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 Fig. 12.
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 (3 cor. 3.) therefore (2. 6. E.) AB is
parallel to CD, or perpendicular to CF, and conse-
quently a tangent to the curve at A (2 cor. 3.); now
BP is a tangent at P, therefore the angle AFB is
equal to BFP (4.), and since the angles FAB, FBP are
right angles, the triangles FAB, FBP are equi-
angular; hence

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 tan-
gent meeting the axis in H, and HK be drawn per-
pendicular to PF, PK shall be equal to half the pa-
rameter of the axis. For the triangles HPK, PFB are
manifestly equiangular, therefore

But if $PD$ be joined, the line $PD$ is evidently perpendicular to the directrix (3.), therefore the figure $HPDF$ is a parallelogram, and $HP=FD$, therefore $PK=FC=\backslash frac\{1\}\{2\}$ the parameter of the axis.

If any ordinate and abscis of a parabola be completed into a parallelogram, the area of the parabola, included between the ordinate and the curve, is two thirds of the parallelogram.

LET $AN$ be any diameter of a parabola, and $PQ$ an ordinate to that diameter. Let $BC$ be drawn through $A$, parallel to $PQ$, and let $PB,\; QC$ be drawn parallel to $NA$; the area comprehended by the curve line $PAQ$ and ordinate $PQ$ is two thirds of the parallelogram $PBCQ$. Join $PA,\; QA$, and draw the tangents $PT,\; QT$, meeting the diameter $NA$ in $T$, and $BC$ in $E$ and $G$; through $E$ and $G$ draw the diameters $EFD,\; GHK$, which will bisect $PA,\; QA$ in $D$ and $K$, (1 cor. 6.), and through $F$ and $H$, the vertices, draw the tangents $RL,\; MV$; join $PF,\; AF$, also $QH,\; AH$. Because $NA=AT$ (7.), and therefore $PQ=2EG$, the triangle $PAQ$ is double the triangle $ETG$. For the same reason the triangles $PFA,\; QHA$ are double the triangles $REL,\; VGM$ respectively, therefore the inscribed figure $PFAHQ$ is double the external figure $TRLMV$. If diameters were drawn through the points $R,\; L,\; M,\; V$, and straight lines were drawn joining the vertices of every two adjacent diameters, also tangents at the vertices of the diameters which pass through the points $R,\; L,\; M,\; V$, there would thus be inscribed in the parabola a new figure which would have the same base $PQ$ as the former, but the number of the remaining sides double that of the former; and corresponding to it there would be a new external figure formed by the tangents at the vertices of the diameters, but still the same proportion between the two figures would hold, or the former would be double the latter, and this would evidently be the case, if the operation of inscribing and circumscribing a new figure were repeated continually. Now it is evident that by thus increasing continually the number of sides of the inscribed figure, it approaches nearer and nearer to the area of the parabola, which is its limit; also that the external figure approaches to the area contained by the two tangents $TP,\; TQ$ and the parabolic arch $PAQ$, which space is its limit; and since the limits of any two quantities which have a constant ratio must have the same proportion to each other as the quantities themselves, the area contained by the parabolic arch $PAQ$ and the ordinate $PQ$ must be double the area contained by the same arch and the two tangents $TP,\; TQ$, and therefore must be two thirds of the area of the triangle $TPQ$, which triangle is evidently equal to the parallelogram $PBCQ$.

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

LET $AB$ be the given directrix, and $F$ the focus. Fig. 13. Place the edge of a ruler $ABKH$ along the directrix $AB$, and keep it fixed in that position. Let $LCG$ be another ruler of such a form that the part $LC$ 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 $AB$ 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.

Through the focus $F$ draw $EFC$ perpendicular to the directrix, and $EC$ will be the axis. Draw any straight line $HEB$ parallel to the directrix, meeting the axis in $E$ any point below the vertex, and on $F$ as a centre, with a radius equal to $CE$, describe a circle cutting $HB$ in $D$ and $d$; these will be points in the parabola required, as is sufficiently evident.

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

LET $DPd$ 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. 6.), 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 (10. & 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. 6.). 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 $A\; B\; a\; b$, 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 Excentricity.

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 Abscissa.

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 $A\; B\; a\; b$ be an ellipse, of which $F,\; f$ are the foci, and $A\; a$ 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 points in the ellipse,

But $D$ and $A$ being points in the ellipse

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 Fig. 17. $Cp$ be not equal to $CP$, take $CQ$ equal to $CP$, and from the points $P,\; q$, $Q$ 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 Fig. 17. $B$, either of the extremities of the conjugate axis.

But because $Aa$ is bisected at $C$,

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 $I$. Because $Df=DG$, and $DI$ is common to the triangles $DfI,\; DGI$ and the angles $fDI,\; GDI$ are equal, these triangles are equal, and $fI=IG$, and hence $fH=HG$ (4. 1. E.), so that $fH+HG=fH+HG$; 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 $KDL$ 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 $KL$, and therefore less than $FD+fD$; therefore the point $K$ will be within the ellipse (2 cor. 1.), and the line $KL$ 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 $I$ falls also at the extremity of the axis, and thus the tangent $DE$, which is always perpendicular to $fI$, 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 $GDE$.

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 shew that if a polished 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, $HPK,\; hPk$ 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 $FPH,\; fpb$ are evidently half the supplements of these angles (4 cor. 4.) therefore the angles $FPH,\; fpb$, and hence $CPH,\; Cpb$ 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 (29. 1. E.).

A straight line drawn from either focus of an ellipse to the intersection of two tangents to the curve, will make equal angles with straight lines drawn from the same focus to the points of contact.

Let $HP,\; Hp$ be tangents to an ellipse at the points $P,\; p$; let a straight line be drawn from $H$, their intersection, to $F$, either of the foci, and let $FP,\; Fp$ be drawn to the points of contact, the lines $PF$ and $pF$ make equal angles with $HF$.

Draw $Pf,\; pf$ to the other focus; in $FP,\; Fp$ produced take $PK=Pf$, and $pk=pF$; join $HK,\; Hk$, and let $fK,\; fk$ be drawn, meeting the tangents at $G$ and $g$. The triangles $fPH,\; KPH$, have $Pf=PK$, by construction, and $PH$ common to both, also the angle $fPH$ equal to $KPH$ (4.), therefore $fH$ is equal to $KH$. In like manner it may be shewn that $fH$ is equal to $kH$, therefore $HK$ is equal to $Hk$; now $FK$ is equal to $Fk$, for each is equal to $FP+Pf$, or $Fp+pF$, that is, to the transverse axis; therefore the triangles $FKH,\; FkH$ are in all respects equal, and hence the angle $KFH$ is equal to $kFH$; therefore $PF$ and $pF$ make equal angles with $HF$.

Two tangents to an ellipse, which are limited by their mutual intersection, and the points in which

which they touch the curve, are to each other reciprocally as the fines of the angles they contain with straight lines drawn from the points of contact to either focus.

Let the straight lines $HP,\; H\backslash rho$, which intersect each other at $H$, be tangents to an ellipse at the points $P,\; \backslash rho$,

Let $KL$ be a triangle, having its base $L$ bisected at $\backslash rho$, and let $Hb$, any straight line parallel to the base, and terminated by the sides, be bisected at $P$; then $P,\; \backslash rho$, 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 $KHPM,\; KL\backslash rho\; N$. The triangles $KHb,\; KL\backslash rho$ being similar, and $Hb,\; L\backslash rho$ similarly divided at $P$ and $\backslash rho$,

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

Next, let $Dd$ meet $K\backslash rho$ in $E$, then

Any straight line not passing through the centre, but terminated both ways by an ellipse, and parallel to a tangent, is bisected by the diameter that passes through the point of contact; or is an ordinate to that diameter.

The straight line $Dd$, terminated by the ellipse, and parallel to the tangent $HPb$, is bisected at $E$, by $P\backslash rho$ the diameter that passes through the point of contact.

Let $L\backslash rho\; l$ be a tangent at the other extremity of the diameter, and let $KD,\; Kd$, tangents at the points $D,\; d$, meet the parallel tangents $HPb,\; L\backslash rho\; l$ in the points $H,\; l$, and draw $DF,\; dF,\; PF,\; pF$ to either focus. Because $Hb$ is parallel to $Dd$,

But, $KD,\; Kd$ being tangents to the ellipse,

therefore (23. 5. E.) fine $bPF\; :\; \backslash text\{fine\; \}\; HDF\; ::\; HD\; :\; bP$;

but fine $HPF$ or fine $bPF\; :\; \backslash text\{fine\; \}\; HDF\; ::\; HD\; :\; HP$, therefore the ratio of $HD$ to $bP$ is the same as that of $HD$ to $HP$, wherefore $PH\; =\; P\backslash rho$. In the same manner it may be demonstrated that $\backslash rho\; L\; =\; \backslash rho\; l$, therefore (Lemma) the diameter $P\backslash rho$ when produced passes

and let $PF,\; pF$ be drawn to either focus; then

Join $HF$, and in $FP$ take $FQ$ equal to $F\backslash rho$, and join $HQ$; then the angles at $F$ being equal (6.) the triangles $HFQ,\; HF\backslash rho$ are equal, therefore $HQ$ is equal to $H\backslash rho$, and the angle $HQF$ is equal to $H\backslash rho\; F$. Now, in the triangle $HPQ$,

through $K$, and bisects $Dd$, which is parallel to $Hb$ or $L\backslash rho$, at $E$.

COR. 1. Straight lines which touch an ellipse at 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. This follows evidently from 2 and 3 cor. to prop. 4. and 2 cor. to prop. 5.

COR. 6. Hence each axis divides the ellipse into two parts which are similar and equal.

If a tangent to an ellipse meet a diameter, and from the point of contact an ordinate be drawn to that diameter, the semi-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 $DK$, a tangent to the curve at $D$, meet the diameter $P\backslash rho$, produced in $K$, and let $DEd$ be an ordinate to that diameter.

Through $P$ and $\backslash rho$, the vertices of the diameter, draw the tangents $PH$ and $\backslash rho\; L$, meeting $KD$ in $H$ and $L$; these tangents are parallel to each other (5.) and to $DE$, the ordinate, by the last proposition. Draw $PF,\; pF$, $DF$ to either of the foci, Then

Now the angles $HPF,\; L\backslash rho\; F$ are equal (cor. 5.) and the fine of $HDF$ is the same as that of $LDF$, therefore

and by alternation,

therefore, because of the parallel lines $PH,\; ED,\; \backslash rho\; L$,

Take

Of the Ellipse. By the same manner of reasoning it may be shewn that because $Bb$ is less than $Aa$,

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.

Fig. 27. Let $DE$ be a semi-ordinate to the diameter $P\backslash rho$, let $PG$ be the parameter of the diameter, and $Qr$ the conjugate diameter. By the definition of the parameter (Def. 13.)

Cor. Let the parameter $PG$ be perpendicular to the diameter $P\backslash rho$; join $\backslash rho\; G$, and from $E$ draw $EM$ parallel to $PG$, meeting $\backslash rho\; G$ in $M$. The square of $DE$, the semi-ordinate, is equal to the rectangle contained by $PE$ and $EM$.

If the rectangles $PGL\backslash rho$, $HGKM$ 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 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.

Fig. 29. Let $P\backslash rho$, $Qr$ 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 $Aa$.

VOL. VI. Part I.

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 $P\backslash rho$, $Qr$ be two conjugate diameters, and $PE$, $Fig.\; 28.$ $QG$ semi-ordinates to any third diameter $Rr$, then $CG^2\; =\; RE\; \backslash cdot\; Er$, 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 $CR^2$ (9.), 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

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 $P\backslash rho$, $Qr$ any two conjugate diameters; draw $PE$, $QG$ perpendicular to $Rr$, and $PL$, $QM$ perpendicular to $Ss$. Then

But $CK\; \backslash cdot\; QM\; =\; \backslash text\{twice\; trian.\; \}\; CKQ\; =\; \backslash text\{paral.\; \}\; CPEQ$, 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 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 semi-diameter 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 semi-diameter to which it is parallel.

LET $PH,\; pb$, tangents at the vertices of a diameter $Pp$ meet $HD,\; b$, a tangent to the curve at any point $D$, in $H$ and $b$; let $CQ$ be the semi-diameter to which the tangents $PH,\; pb$ are parallel, and $CR$ that to which $Hb$ is parallel, then,

If the tangent $HD,\; b$ be parallel to $Pp$ the proposition is manifest. If it is 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 $PH,\; ED,\; CK,\; pb$,

Again, the triangles $LED,\; CMR$ are evidently similar, and $LE,\; LD$ similarly divided at $H$ and $P$, also at $b$ 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. 12.) and therefore $PE\; \backslash cdot\; pE\; =\; CM^2$ (14.)

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 semi-diameter to which the tangent is parallel.

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

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 $DP,\; d$ 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. Of the Ellipse. Fig. 31.

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 $FDP,\; EPD$ equal (4.) and the side $DP$ common to both, they are therefore equal, and consequently have $ED\; =\; DF$, and $EP\; =\; PF$, wherefore $Ef\; =\; FP\; +\; Pf\; =\; Aa$. 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,

but $FC$ is half of $Ff$, therefore $CD$ is half of $Aa$. In like manner it may be shewn 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 $Cd,\; PG,\; CD,\; Pg$ 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.

LET $DP,\; d$ 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. Fig. 32.

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 $FD,\; fd$ 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 $fP,\; FDP$ are right angles, therefore the triangles $fPd,\; FPD$ are similar, and

If from $C$ the centre of an ellipse a straight line $Fig.\; 33.$ $CL$ 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 $b$, 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 $Db$ 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 $DE,\; Dd$, which will be perpendicular to the axes. The

The triangles DEH, CLm are evidently equiangular, therefore

In the same way it is shewn 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

But $GD\; =\; AC$ (cor. 17.) 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.).

Fig. 33. 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 $HGb$, drawn through $G$ perpendicular to the axis, is called the Directrix of the ellipse.

Cor. 1. If $MFm$, 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 $G$. (9.)

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 semi-transverse axis to the eccentricity.

Fig. 33. 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 $L\; \backslash cdot\; a\; =\; Df$, 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

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, and from similar triangles,

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

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 $ai\; =\; aF$. This follows evidently from last corollary.

Let $Aa,\; Bb$ be the transverse and conjugate axes Fig. 34 of an ellipse; from $K$ any point in the conjugate axis let a straight line $KH$, which is equal to the sum or difference of the semi-axes $CA,\; CB$, be placed so as to meet the transverse axis in $H$, and in $KH$, produced beyond $H$ when $KH$ is the difference of the semi-axes, let $HD$ be taken equal to $CB$; the point $D$ is in the ellipse.

DRAW $DE$ perpendicular to $Aa$, and through $C$ draw $CG$ parallel to $KD$, meeting $ED$ in $G$, then $CG\; =\; KD\; =\; AC$ by construction, hence $G$ is in the circumference of a circle of which $C$ is the centre, and $CA$ the radius; and because the triangles $CEG,\; HED$ are similar,

therefore $DE$ is a semi-ordinate, and $D$ a point in the ellipse (5 Cor. 11.).

The instrument called the trammel, also the elliptic compass, which workmen use for describing elliptic curves, is constructed on the property of the curve demonstrated in this proposition. (See COMPASSES.) Upon the same principle lathes are constructed for turning picture frames, &c. of an oval form.

If a circle be described on the transverse axis of an ellipse as a diameter, the area of the circle will be to the area of the ellipse as the transverse axis to the conjugate axis.

Let $Aa$ be the transverse axis of the ellipse, which Fig. 15. is also the diameter of the circle. Draw $DE,\; D\text{'}E\text{'},\; D\text{'}\text{'}E\text{'}\text{'}$ any number of perpendiculars to the axis, meeting the ellipse in $D,\; D\text{'},\; D\text{'}\text{'}$, and the circle in $d,\; d\text{'},\; d\text{'}\text{'}$, and join $AD,\; DD\text{'},\; D\text{'}\text{'}D\text{'}\text{'}$; also $Aa,\; d,\; d\text{'},\; d\text{'}\text{'}$,
3 X 2

Of the
Hyperbola.

$d\text{'}d\text{'}$, $d\text{'}a$, and draw $DG$, $dg$ parallel to $Aa$, meeting $d\text{'}E\text{'}$ in $G$ and $g$.

The triangle $A\; d\; E$ is to the triangle $ADE$ as $d\text{'}E$ to $DE$, that is (4. cor. 11.) as the transverse axis to the conjugate axis. Again, because $d\text{'}E\text{'}$ and $dE$ are similarly divided at $D\text{'}$ and $D$ (3. cor. 11.)

But, triangle $dg\; d\text{'}$ : triangle $DGD\text{'}$ :: $dg\; :\; DG$, therefore the triangles $dg\; d\text{'}$, $DGD\text{'}$ as well as the rectangles $d\text{'}E\text{'}$, $DE\text{'}$ are to each other as $dE$ to $DE$, or as the transverse axis to the conjugate axis, and consequently the trapezoids $dEE\text{'}d\text{'}$, $DEED\text{'}$ are to each other in the same ratio. In like manner it may be shewn, that the trapezoids $d\text{'}E\text{'}E\text{'}\text{'}d\text{'}$, $D\text{'}E\text{'}\text{'}ED\text{'}$, also the triangles $d\text{'}E\text{'}\text{'}a$, $D\text{'}E\text{'}\text{'}a$ are to each other as the transverse to the conjugate axis, and therefore the whole rectilinear figure $A\; d\; d\text{'}\; d\text{'}\text{'}\; a$ inscribed in the semicircle to the whole figure $ADD\text{'}D\text{'}\text{'}a$ inscribed in the semiellipse in the same ratio, which ratio is constant, and altogether independent of the number of the sides of each figure. But, the base $Aa$ remaining common to both figures, if we suppose the number of perpendiculars $dDE$, $d\text{'}E\text{'}$, &c. indefinitely increased, it is evident that the polygons $A\; d\; d\text{'}\; d\text{'}\text{'}\; a$, $ADD\text{'}D\text{'}\text{'}a$ will approach nearer and nearer to the semicircle and semiellipse, which are their respective limits, therefore, the semicircle is to the semiellipse, and consequently the circle is to the ellipse, as the transverse to the conjugate axis.

COR. The area of an ellipse is equal to the area of a circle, whose diameter is a mean proportional between the axes.

PROP. XXIII. PROBLEM.

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.

FIRST METHOD. By a Mechanical Description.

Fig. 36.

LET $Aa$, $Bb$ be the transverse and conjugate axes, and $C$ the centre. On $B$, one extremity of the conjugate axis, as a centre, with a radius equal to $AC$, half the transverse axis, let a circle be described, cutting the transverse axis in $F$ and $f$; these points will be the foci of the ellipse (4. cor. 1.).

PART III. OF THE HYPERBOLA.

DEFINITIONS.

Fig. 39.

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 $DAD\text{'}$ 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 $DAD\text{'}$, $da\; d\text{'}$, de-

scribed, opposite to one another, 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. A straight line passing through the centre, and terminated

SECOND METHOD. By Finding any Number of Points in the Curve.

Find $F$ either of the foci as before; draw $HAK$, Fig. 37. $hak$, perpendicular to the transverse axis at its extremities, and take $AH$ and $AK$ on each side of the vertex equal to $AF$, also $ab$ and $ak$ each equal to $aF$, join $Hb$ and $Kk$, take $E$ any point in $Aa$, and through $E$ draw $NE$ parallel to $HK$, meeting $Hb$ and $Kk$ in $N$ and $n$. On $F$ as a centre with a radius equal to $EN$ or $En$ let a circle be described, meeting $Nn$ in $D$ and $d$, these will be two points in the ellipse, and in the same way may any number of points be found. The reason of this construction is obvious from cor. 1. and 2. to prop. 20.

PROP. XXIV. PROBLEM.

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

LET $ABab$ be the given ellipse. Draw two parallel chords $Hb$, $Kk$, 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. 8.). 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 fall wholly without the curve, then $CD$ must be half the transverse axis; and if it fall wholly within the curve, then $CD$ must be half the conjugate axis (12.). If the circle neither falls wholly without the curve, nor wholly within it, let the circle meet it again in $d$, join $Dd$, and bisect $Dd$ in $E$, join $CE$, which produce to meet the ellipse in $A$ and $a$, then $Aa$ will be one of the axes (5. cor. 8.), for it is perpendicular to $Dd$ (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 $Aa$.

scribed, opposite to one another, 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. A straight line passing through the centre, and terminated

Of the Hyperbola. terminated by the opposite hyperbolas, is called a Transverse Diameter. It is also sometimes called, simply, a Diameter.

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

VI. 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 take place when A is between F and f.

VII. 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.

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

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

X. 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.

Fig. 39. LET DAD', d'a'd' be opposite hyperbolas, of which F, f are the foci, and A a the transverse axis; let D be any point in the curve, and DF, Df lines drawn to the foci, Df—DF=Aa.

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.

Fig. 40. 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 ne-

cessarily 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. Let C be the centre; then CF=Cf (Def. 3.), and FA=fa, therefore CA=Ca.

Two triangles ABC, ADC on the same base, and Fig. 41. on the same side of it, having AB, AD, the greater of the two sides of each ending in the same extremity of the base, and having their vertical angles B, D without each other, cannot have the difference of the sides of the one equal to the difference of the sides of the other.

LET AD meet BC in E. Because AE+EB is greater than AB, (AE+EB)—BC=AE—EC is greater than AB—BC. Again, because DC is less than DE+EC, AD—DC is greater than AD—(DE+EC)=AE—EC; much more therefore is AD—DC greater than AB—BC. Therefore AD—DC cannot be equal to AB—BC.

Every transverse diameter of an hyperbola is bisected in the centre. Plate CLIX.

LET Pp be a transverse diameter, it is bisected in Fig. 42. C; for if Cp be not equal to CP, take CQ equal to CP; from the points P, p, Q draw straight lines to F and f the foci; draw fD perpendicular to Cp, and FE parallel to PD, meeting fD in E; join E, p, EQ. Because fC=CF, and CD is parallel to EF, fD=DE (2. 6. E.). Now pD is common to the triangles fDp, EDp, and the angles at D are equal, being right angles, therefore the triangles are equal, and pf=pE. In like manner it appears that Qf=QE. Again, 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 Pf=QF, therefore FO—fQ is equal to fP—FP, or (Def. 1.) to Fp—fp; that is, FO—QE is equal to Fp—pE, which by the preceding lemma is absurd; therefore CP=Cp.

COR. 1. Every transverse diameter meets the opposite hyperbolas each in one point only, and being produced falls within them.

COR. 2. Every transverse diameter divides the opposite hyperbolas into parts which are equal and similar; the like parts of the curve being at opposite extremities of the diameter, and on contrary sides of it.

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.

Of the
Hyperbola.
Plate
CLVIII.
Fig. 39.

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

But because $\backslash Delta\; a$ is bisected at $C$, and produced to $f$,

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.

Plate
CLIX.
Fig. 43.

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 $fDF$ is a tangent to the curve.

Take $H$ any other point in $DE$, take $DG\; =\; Df$, and join $Hf$, $HF$, $HG$, $fG$; let $fG$ meet $DE$ in $I$. Because $Df\; =\; DG$ and $DI$ is common to the triangles $DfI$, $DGI$, and the angles $fDI$, $GDI$ are equal, these triangles are equal, and $fI\; =\; IG$, and hence $fH\; =\; HG$ (4. 1. E.), so that $fH\; =\; fH\; =\; FH\; =\; HG$; 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 $DHI$ 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 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 $I$ falls also at the extremity of the axis, and thus the tangent $DE$, which is always perpendicular to $fI$, 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 $DI$ meet the axis in $E$. Because $DE$ bisects the angle $FDf$,

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

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 $P\backslash rho$ be a diameter, $HP$, $b\backslash rho$ tangents at its vertices; draw straight lines from $P$ and $\backslash rho$ to $F$ and $f$ the foci. The triangles $FCP$, $fC\backslash rho$, having $FC\; =\; fC$, $CP\; =\; C\backslash rho$ (2.), and the angles at $C$ equal, are in all respects equal, and because the angle $FPC$ is equal to $C\backslash rho\; f$, $FP$ is parallel to $f\backslash rho$ (27. 1. E.), therefore $Pf$ is equal and parallel to $\backslash rho\; F$ (33. 1. E.): thus $FPf\backslash rho$ is a parallelogram of which the opposite angles $P$ and $\backslash rho$ are equal (34. 1. E.); now the angles $FPH$, $f\backslash rho\; b$ are the halves of these angles (4.); therefore the angles $FPH$, $f\backslash rho\; b$, and hence $CPH$, $C\backslash rho\; b$, are also equal, and consequently $HP$ is parallel to $b\backslash rho$.

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 $F\backslash rho\; b$ is equal to $FPH$.

COR. 2. The transverse axis is the only diameter which is perpendicular to tangents at its vertices. For let $P\backslash rho$ 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.

A straight line drawn from either focus of an hyperbola to the intersection of two tangents to the curve, will make equal angles with straight lines drawn from the same focus to the points of contact.

Let $HP$, $H\backslash rho$ be tangents to an hyperbola at the points $P$, $\backslash rho$; let a straight line be drawn from $H$ their intersection to $F$ either of the foci; and let $FP$, $f\backslash rho$ be drawn to the points of contact; the lines $PF$, $\backslash rho\; F$ make equal angles with $HF$.

Draw $Pf$, $\backslash rho\; f$ to the other focus. In $PF$ and $\backslash rho\; F$ take $PK\; =\; Pf$, and $\backslash rho\; k\; =\; \backslash rho\; f$; join $HK$, $Hk$, and let $fK$, $f\; k$ be drawn, meeting the tangents in $G$ and $g$. The triangles $fPH$, $KPH$ have $Pf\; =\; PK$, by construction, and $PH$ common to both, also the angle $fPH$ equal to $KPH$ (4.); therefore $fH$ is equal to $KH$. In like manner it may be shewn that $fH$ is equal to $kH$, therefore $HK$ is equal to $Hk$; now $FK$ is equal to $fK$, for each is equal to the difference between $FP$ and $Pf$, or $F\backslash rho$ and $f\backslash rho$, that is, to the transverse axis; therefore the triangles $FKH$, $f\; k\; H$ are in all respects equal, and hence the angle $KFH$ is equal to $kFH$, therefore $PF$ and $\backslash rho\; F$ make equal angles with $HF$.

Two tangents to an hyperbola, or opposite hyperbolas, which are limited by their mutual intersection and the points in which they touch the curve, are to each other, reciprocally, as the sines of the angles they contain with straight lines drawn from the points of contact to either focus.

Fig. 47 and 48. Let $HP,\; H\backslash rho$, which intersect each other at $H$, be tangents to an hyperbola, or opposite hyperbolas, at the points $P,\; \backslash rho$; and let $PF,\; \backslash rho\; F$ be drawn to either focus,

Join $HF$, and in $FP$ take $FQ$ equal to $F\backslash rho$, and join $HQ$; then, the angles at $F$ being equal (6.), the triangles $HFO,\; HF\backslash rho$ are equal, therefore $HQ$ is equal to $H\backslash rho$, and the angle $HQF$ is equal to $H\backslash rho\; F$. Now in the triangle $HPQ$,

Fig. 49. Let $KL$ be a triangle, having its base $L$ bisected at $\backslash rho$, and let $Hb$, any straight line parallel to the base, and terminated by the sides produced, be bisected at $P$, then $P,\; \backslash rho$ 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 $KP,\; K\backslash rho$. The triangles $KHb,\; KL$ being similar, and $Hb,\; L$ similarly divided at $P,\; \backslash rho$,

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

Next let $Dd$ meet $K\backslash rho$ in $E$, then

Any straight line terminated both ways by an hyperbola, and parallel to a tangent, is bisected by the transverse diameter produced, that passes through the point of contact, or is an ordinate to that diameter.

Fig. 50. The straight line $Dd$, terminated by the hyperbola, and parallel to the tangent $HPb$, is bisected at $E$ by $P\backslash rho$ the transverse diameter produced, which passes through $P$, the point of contact.

Let $L\backslash rho$ be a tangent at the other extremity of the diameter, and let $KD,\; Kd$, tangents at the points $D,\; d$, meet the parallel tangents $HPb,\; L\backslash rho$ in the

points $H,\; L,\; b,\; l$, and draw $DF,\; dF,\; PF$ to either focus. Because $Hb$ is parallel to $Dd$,

But $KD,\; Kd$ being tangents to the hyperbola,

therefore $\backslash text\{sine\; \}\; bdF\; :\; \backslash text\{sine\; \}\; HDF\; ::\; HD\; :\; bd$,

now, $\backslash text\{sine\; \}\; bPF\; :\; \backslash text\{sine\; \}\; bdF\; ::\; bd\; :\; bP$ (7.)

therefore, (23. 5. E.) $\backslash text\{sine\; \}\; bPF\; :\; \backslash text\{sine\; \}\; HDF\; ::\; HD\; :\; bP$,

but $\backslash text\{sine\; \}\; HPF$ or $\backslash text\{sine\; \}\; bPF\; :\; \backslash text\{sine\; \}\; HDF\; ::\; HD\; :\; HP$, therefore the ratio of $HD$ to $bP$ is the same as the ratio of $HD$ to $HP$, therefore $PH\; =\; Pb$. In the same manner it may be demonstrated that $\backslash rho\; L\; =\; \backslash rho\; l$, therefore (lemma 2.) the diameter $P\backslash rho$ when produced passes through $K$, and bisects $Dd$, which is parallel to $Hb$, or $Ll$, at $E$.

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

COR. 2. Every ordinate to a transverse 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 transverse diameter are parallel to each other.

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

COR. 5. The ordinates to the transverse axis are perpendicular to it, and no other transverse diameter has its ordinates perpendicular to it. This follows from 2. cor. 4. and 2. cor. 5.

COR. 6. The transverse axis, indefinitely produced, divides each of the opposite hyperbolas into two parts which are similar to one another.

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.

Let $DK$ a tangent to the curve at $D$ meet the transverse diameter $P\backslash rho$ in $K$, and let $DEd$ be an ordinate to that diameter,

Through $P$ and $\backslash rho$, the vertices of the diameter, draw the tangents $PH$ and $\backslash rho\; L$, meeting $KD$ in $H$ and $L$, these tangents are parallel to each other (5.), and to $DE$, the ordinate, by last proposition. Draw $PF,\; \backslash rho\; F,\; DF$ to either of the foci. Then,

Now the angles $HPF,\; L\backslash rho\; F$ are equal (1. cor. 5.); therefore,

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.

Fig. 51. 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

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 $FDF$ 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 $LGH$, therefore,

Fig. 52. XI. If through $A$, one of the vertices of the transverse axis, a straight line $HA\; b$ be drawn, equal and parallel to $Bb$ the conjugate axis, and bisected at $A$ by the transverse axis, the straight lines $CHM,\; Cb\; m$ 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 $H\text{'}a\; b\text{'}$, parallel to $Bb$, and meeting the asymptotes of the hyperbola $DAD$ in $H\text{'}$ and $b\text{'}$. Because $aC$ is equal to $AC$, $aH\text{'}$ is equal to

Cor. 1. The rectangle contained by $PE$ and $E\backslash rho$ is equal to the rectangle contained by $KE$ and $CE$.

Cor. 2. The rectangle contained by $PK$ and $K\backslash rho$ is equal to the rectangle contained by $KE$ and $KC$.

$A\; b$, or to $BC$; also $a\; b\text{'}$ is equal to $AH$, or to $BC$; hence, by the definition, $CH\text{'}$ and $Cb\text{'}$ are asymptotes of the opposite hyperbola $da\; d$.

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 passing through $B$ parallel to $Aa$; the same is true of the points $b,\; b\text{'},\; b\text{'}$.

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 that 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 $MD\; \backslash cdot\; Dm$ is 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 $CE$ (9.), and therefore

Again, $CB$ being a mean proportional between $CL$ and $CG$ (10.)

consequently $EM^2$ is greater than $ED^2$, and $EM$ greater

Of the Hyperbola. 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 hyperbola. Again, the straight line Mm terminated by the asymptotes, being manifestly bisected by the axis at E,

but it has been shewn that

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 · 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 · Dm, NR · 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.

Fig. 53. 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.

Because $NO^2$ is greater than $ME^2$, and $NR\; \backslash cdot\; Rn\; =\; MD\; \backslash cdot\; Dm$, (2. cor. 11.) therefore $NO^2\; -\; NR\; \backslash cdot\; Rn$ is greater than $ME^2\; -\; MD\; \backslash cdot\; Dm$, that is $RO^2$ is greater than $DE^2$, and $RO$ is greater than $DE$; now $On$ is greater than $Em$, therefore $Rn$ is greater than $Dm$, and since $Rn\; :\; Dm\; ::\; DM\; :\; RN$, (2. Cor. 11.) $DM$ is greater than $RN$,

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 $Ab$, 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.

VOL. VI. Part II.

The name asymptotes (non concurrentes) has been given to the lines CH, Cb, 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 a 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 Fig. 54. and opposite hyperbolas, let parallel lines $EDe$, $HGb$ be drawn to meet the asymptotes in E, e, and H, b, 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 $bGm$, $eDl$,

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

COR. 1. If a straight line be drawn through D, Fig. 54. and 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 $ED\; \backslash cdot\; De$, $HG\; \backslash cdot\; Gh$ have been proved to be equal, it may be shewn that the rectangles $Ed\; \backslash cdot\; de$, $HG\; \backslash cdot\; Gb$ 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 $Ee$ will thus become a tangent to the curve at P. Therefore any tangent $KP$, which is terminated by the asymptotes, is bisected at P, the point of contact.

COR. 3. And if any straight line $KP$, limited by the asymptotes, be bisected at P a point in the curve, that line is 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, Fig. 54. any point in the hyperbola, parallel to a tangent $KP$, 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 an 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 $db$ and $dk$ parallel to DH and DK. The rectangles $HD\; \backslash cdot\; DK$, $bd\; \backslash cdot\; dk$ are equal.

Join D, $d$ meeting the asymptotes in E, e. From similar triangles

therefore, taking the rectangles of corresponding terms,

COR. 1. If the lines $D\text{'}K\text{'}$, $D\text{'}H\text{'}$, $d\text{'}k\text{'}$, $d\text{'}h\text{'}$, be parallel to the asymptotes, and thus form the parallelograms $D\text{'}K\text{'}CH\text{'}$, $d\text{'}k\text{'}CB\text{'}$, 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 $D\text{'}K\text{'}C$, $d\text{'}k\text{'}C$ are also equal.

COR. 2. If from $D\text{'}$, $d\text{'}$, any two points in an hyperbola, straight lines $D\text{'}K\text{'}$, $d\text{'}k\text{'}$ 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 cor. and 14. 6. E.

XII. If $Aa$ be the transverse axis, and $Bb$ the conjugate axis of two opposite hyperbolas $DAD$, $d\; a\; d$, and if $Bb$ be the transverse axis, and $Aa$ the conjugate axis of other two opposite hyperbolas $EBE$, $e\; b\; e$, 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 $DAD$, $d\; a\; d$ are also the asymptotes of the hyperbolas $EBE$, $e\; b\; e$. 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 ^{Of the} an ordinate to that diameter. ^Hyperbola.

THE straight line $Dd$ terminated by the opposite Fig. 58. hyperbolas, and parallel to the tangent $KQk$, is bisected at E by $Qg$ 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. 13.) therefore $GE\; =\; Eg$; now $DG\; =\; gd$ (1 cor. 13.), 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.

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

COR. 4. A straight line that bisects two parallel straight lines which terminate in the opposite hyperbolas is a second diameter.

COR. 5. The ordinates to the conjugate or second axis are perpendicular to it, and no other second diameter is perpendicular to its ordinates.

COR. 6. 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 Fig. 59. parallel to $DEd$, any ordinate to the second diameter $Qg$, the second diameter $Qg$ shall be parallel to the ordinates to the diameter $Pp$.

Draw the diameter $dCG$ through one extremity of the ordinate $dD$, and join G and D, the other extremity, meeting $Pp$ in H. Because $dG$ is bisected at C, and $CH$ is parallel to $dD$, the line $DG$ is bisected at H, therefore $DG$ is an ordinate to the diameter $Pp$. And because $dG$ and $dD$ are bisected at C and E, the diameter $Qg$ is parallel to $DG$ (2. 6. E.), therefore $Qg$ 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 Lotus 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 $PCp$ be any transverse diameter of an hyper- Fig. 60. bola, $HPb$ a tangent at its vertex, meeting the asymptotes

Of the Hyperbola: totes in $H$ and $b$, and $Qg$ the diameter which is conjugate to $P\backslash rho$; the tangent $Hb$ is equal to the diameter $Qg$.

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 $E$ and $e$, 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 $DEK$, $dEL$ are similar, as also $eDk$, $eDl$, therefore

therefore, taking the rectangles of the corresponding terms,

But $KD\; \backslash cdot\; Dk\; =\; BC^2$ (11.) and $BC^2\; =\; Ld\; \backslash cdot\; dl$ (5 cor. 13.) therefore $ED\; \backslash cdot\; De\; =\; Ed\; \backslash cdot\; de$.

Now $ED\; \backslash cdot\; De\; =\; HP^2$ (4 cor. 13.), and $Ed\; \backslash cdot\; de\; =\; QC^2$ (5 cor. 13.) therefore $HP^2\; =\; QC^2$, and $HP\; =\; QC$, and consequently $Hb\; =\; Qg$.

COR. 1. If another tangent be drawn to the curve at $\backslash rho$, meeting the asymptotes in $H\text{'}$ and $b\text{'}$, the straight lines which join the points $H$, $H\text{'}$, also $b$, $b\text{'}$, are tangents to the conjugate hyperbolas at $Q$ and $q$. For $\backslash rho\; H\text{'}$ as well as $\backslash rho\; H$ is equal and parallel to $CQ$, therefore the points $H$, $Q$, $H\text{'}$ are in a straight line parallel to $P\backslash rho$, and $HQ\; =\; H\text{'}Q$ (33. 1. E), therefore $HQH\text{'}$ is a tangent to the curve at $Q$. In like manner it appears that $bqb\text{'}$ 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 a 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 $Qg$ in $L$, and let $DGd\text{'}$ be an ordinate to that diameter, then

Let $P\backslash rho$ be the diameter that is conjugate to $Qg$, let $HPb$ be a tangent at the vertex, terminated by the asymptotes; through $D$ draw the ordinate $DEd\text{'}$ to the diameter $P\backslash rho$, meeting the asymptotes in $M$ and $m$; let $K$ be the intersection of $DL$ and $P\backslash rho$. Because $DK$ is a tangent at $D$, and $DEd\text{'}$ an ordinate to $P\backslash rho$, $CP$ is a mean proportional between $CE$ and $CK$ (9.) and therefore

Now, the lines $CQ$, $PH$, $EM$ being parallel (8. and Def. 15.), from similar triangles,

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

Fig. 62. LET $DEd\text{'}$ be an ordinate to the transverse diameter $P\backslash rho$, and let $Qg$ be its conjugate diameter,

Let $DKL$ a tangent at $D$ meet the diameter in $K$, and its conjugate in $L$. Draw $DG$ parallel to $P\backslash rho$, meeting $Qg$ in $G$. Because $CP$ is a mean proportional between $CE$ and $CK$ (9.)

and because $CQ$ is a mean proportional between $CG$ and $CL$ (18.)

and by inversion and alternation,

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 $Qg$. It has been shewn 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 abscissæ; and the squares of semi-ordinates, and of ordinates,

nates 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.

Plate CLXI. 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, $P\backslash rho$ 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 $P\backslash rho$ is greater than $Rr$. In like manner it is shewn that if $Ss$ be the conjugate axis, and $Qq$ any other second diameter, $Qq$ is greater than $Ss$.

Plate CLXI. If an ordinate be drawn to any transverse diameter of an hyperbola, the rectangle under the abscissæ 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 $P\backslash rho$; let $PG$ be the parameter of the diameter, and $Qq$ the conjugate diameter. By the definition of the parameter (Def. 16.)

$P\backslash rho\; :\; Qq\; ::\; Qq\; :\; PG$,
therefore $P\backslash rho\; :\; PG\; ::\; P\backslash rho^2\; :\; Qq^2$, (2 cor. 20. 6. E.)
But $P\backslash rho^2\; :\; Qq^2\; ::\; PE\; \backslash cdot\; E\backslash rho\; :\; DE^2$, (19.)
therefore $PE\; \backslash cdot\; E\backslash rho\; :\; DE^2\; ::\; P\backslash rho\; :\; PG$.

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

For $PE\; \backslash cdot\; E\backslash rho\; :\; DE^2\; ::\; P\backslash rho\; :\; PG$,
and $P\backslash rho\; :\; PG\; ::\; E\backslash rho\; :\; EM\; ::\; PE\; \backslash cdot\; E\backslash rho\; :\; PE\; \backslash cdot\; EM$,
therefore $DE^2\; =\; PE\; \backslash cdot\; EM$.

If the rectangles $PGL\backslash rho$, $HGKM$ 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 abscissæ $PE$, 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 square of the ordinate above the rectangle contained by the abscissæ and parameter that Apollonius gave the curve to which the property belonged the name of Hyperbola.

Plate CLXI. 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 $P\backslash rho$, $Qq$ be two conjugate diameters, of which $P\backslash rho$ 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 $KC\; \backslash cdot\; CG$ are equal, for each is equal to $CR^2$ (9.) therefore,

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

$HC\; :\; CK\; ::\; HE\; :\; CG$,
wherefore $CG\; :\; CE\; ::\; HE\; :\; CG$,
consequently $CG^2\; =\; CE\; \backslash cdot\; EH\; =\; (1\; \backslash text\{\; cor.\; 9.\})\; RE\; \backslash cdot\; Er$.

Again, from the similar triangles $HPC$, $CQK$,

Now it was shewn that $HC\; :\; CK\; ::\; CG\; :\; CE$,

therefore $CG\; :\; CE\; ::\; CE\; :\; KG$,
consequently
$CE^2\; =\; CG\; \backslash cdot\; GK\; =\; (3.\; 2.\; E.)\; CG^2\; +\; GC\; \backslash cdot\; CK$.
But $GC\; \backslash cdot\; CK\; =\; CR^2$ (18.)
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 shewn that $CE^2\; =\; CG^2\; +\; CR^2$;

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 $P\backslash rho$, $Qq$ 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 $P\backslash rho^2\; -\; Qq^2\; =\; Rr^2\; -\; Ss^2$.

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 $P\backslash rho,\; Q\backslash sigma$ 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 $CK\; :\; CA\; ::\; CL\; :\; CL$ (9.)
and $CA\; :\; CB\; ::\; CL\; :\; QM$ (1 cor. 22.)
ex aeq. $CK\; :\; CB\; ::\; CA\; :\; QM$,
therefore $CK\; \backslash cdot\; QM\; =\; CB\; \backslash cdot\; CA$.

But $CK\; \backslash cdot\; QM\; =$ twice tria. $CKQ\; =$ paral. $CPEQ$,
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 semi-diameter 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 semi-diameter to which it is parallel.

Let $PH,\; \backslash rho\; b$, tangents at the vertices of a transverse diameter $P\backslash rho$, meet $DH,\; b$, a tangent to the curve at any point $D$, in $H$ and $b$; let $CQ$ be the semi-diameter to which the tangents $PH,\; \backslash rho\; b$ are parallel, and $CR$ that to which $DH$ is parallel; then

Let $Hb$ meet the semi-diameters $CP,\; CQ$ in $L$ and $K$. Draw $DE,\; RM$ parallel to $CQ$ and $DG$ parallel to $CP$.

Because $LP\; \backslash cdot\; L\backslash rho\; =\; LE\; \backslash cdot\; LC$ (2 cor. 9.)
$LP\; :\; LE\; ::\; LC\; :\; L\backslash rho$;

hence, and because of the parallels $PH,\; ED,\; CK,\; \backslash rho\; b$,

wherefore $PH\; \backslash cdot\; \backslash rho\; b\; =\; ED\; \backslash cdot\; CK$.

But $ED\; \backslash cdot\; CK\; =\; CG\; \backslash cdot\; CK\; =\; CQ^2$ (18.)
therefore $PH\; \backslash cdot\; \backslash rho\; b\; =\; CQ^2$.

Again, the triangles $LED,\; CMR$ are evidently similar, and $LE,\; LD$ are similarly divided at $H$ and $P$, also at $b$ and $\backslash rho$.

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. 15.) and therefore $PE\; \backslash cdot\; \backslash rho\; E\; =\; CM^2$ (22.)

COR. The rectangle contained by $LD$ and $DK$, the segments of a tangent intercepted between $D$ the point of contact, and $P\backslash rho,\; Q\backslash sigma$, any two conjugate diameters, is equal to the square of $CR$, the semi-diameter to which the tangent is parallel.

Let the parallel tangents $PH,\; \backslash rho\; b$ meet $LK$ in $H$ and $b$, and draw $DE$ a semi-ordinate to $P\backslash rho$. Because of the parallels $ED,\; PH,\; CK,\; \backslash rho\; b$,

therefore

But $LE\; \backslash cdot\; EC\; =\; EP\; \backslash cdot\; E\backslash rho$ (1 cor. 9.)
therefore $LD\; \backslash cdot\; DK\; =\; DH\; \backslash cdot\; Db\; =$ (by this prop.) $CR^2$.

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 $FP,\; fP$, 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 (4.) and the side $DP$ common to both; they are therefore equal, and consequently have $ED\; =\; DF$, and $EP\; =\; PF$, wherefore $Ef\; =\; FP\; -\; Pf\; =\; Aa$. 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 $Ff$, therefore $CD$ is half of $Aa$.

COR. If a straight line $Qg$ 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 $Pg\; =\; DC\; =\; AC$.

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.

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 $FD$ 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

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 $FD,\; fP$ are right angles, therefore the triangles $FD,\; fP$ are similar, and

Fig. 64. 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 $b$, 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.

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

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

In the same way it may be shewn 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

Fig. 69. 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 semi-transverse axis, a straight line $HGb$ drawn through $G$, perpendicular to the axis, is called the Directrix of the hyperbola.

COR. 1. If $MF$ 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 $G$ (9.).

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 semi-transverse 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 $L\; a\; =\; Df$; 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

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 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 semi-diameters 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 $G$. The triangles $CDP,\; CEQ$ are equal (1 cor. 14.) therefore, taking the triangle $CDG$ from both, the triangle $CGP$ is equal to the quadrilateral $DEQG$; to these add the figure $PGQ$.

Of the Hyperbola. $PGQ$ and the hyperbolic sector $PCQ$ is equal to the hyperbolic trapezium $PDEQ$.

Fig. 71. 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.

THROUGH $Q$ draw a tangent to the curve, meeting the asymptotes in $\backslash kappa$ and $L$; join $PR$ meeting the asymptotes in $M$ and $N$; draw the semi-diameters $CP$, $CQ$, $CR$, let $CQ$ meet $PR$ in $G$.

Because $QE$ is parallel to $CM$, and $KQ$ is equal to $QL$ (2 cor. 13.) $CE$ is equal to $EL$; and because $MC$, $PD$, $RH$, are parallel, and $MP$ is equal to $RN$ (1 cor. 13.) $CD$ is equal to $HN$. Now, by hypothesis,

$CD\; :\; CE\; ::\; CE\; :\; CH$,
therefore $NH\; :\; LE\; ::\; CE\; :\; CH$;
but $CE\; :\; CH\; ::\; HR\; :\; EQ$ (2 cor. 14.)
therefore $NH\; :\; LE\; ::\; HR\; :\; EQ$,
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 $CQ$ (8.) 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 $PGQ$, $RQG$ be taken from the equal triangles $PCG$, $RCG$, and there will remain the hyperbolic sectors $PCQ$, $RCQ$ equal to each other. Therefore (29.) 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.

Fig. 72, 73. 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 axis.

FIRST METHOD. By a Mechanical Description.

Fig. 72. JOIN $AB$, and in $A$ $a$, 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 now 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, $Df\; =\; DF$ is by hypothesis equal to $Aa$ the transverse axis.

SECOND METHOD. By finding any number of points in the Curve.

Find $F$, either of the foci as before, draw $HAK$, Fig. 73-
$ba$ perpendicular to the transverse axis at its extremities, and $AH$ and $AK$ on each side of the vertex equal to $AF$, also $ab$ and $ak$ each equal to $aF$; join $Hb$ and $Kk$; take $E$ any point in $Aa$, and though $E$ draw $NE$ parallel to $HK$, meeting $Hb$ and $Kk$ in $N$ and $n$. On $F$ as a centre, with a radius equal to $EN$ or $En$, let a circle be described meeting $Nn$ in $D$ and $d$, these will be two points in the hyperbola; and in the same way may any number of points in the hyperbola, or opposite hyperbolas, be found. The reason of this construction is obvious from cor. 1. and 2. to Prop. 28.

An hyperbola being given by position, to find its axes. Plate CLXII.

Let $HA$ be the given hyperbola. Draw two parallel straight lines $Hb$, $Kk$ 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. 8.) 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 (20.), 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. 8.) 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 $CB$ to 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 (19.).

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 is 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 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 plane 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

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 Fig. 76. 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 HFKM 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 EFGM be a section parallel to the base, passing through EG; then the two planes HFKM, EFGM being perpendicular to the plane VAB, their common section FLM is perpendicular to ELG, and since EFGM 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 HFKM 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 section 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 Fig. 77, 78. common section of a plane passing through its vertex 79. 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

Of the Cone the other side in H, fig. 77. within the cone; it will be parallel to VA in fig. 78, and it will meet VA, produced beyond the vertex, in H, fig. 79.

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 $FL^2\; =\; EL\; \backslash cdot\; LG$.

Fig. 77. CASE I. 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

Now the ratio of $KH^2$ to $HQ\; \backslash cdot\; KR$ is the same wherever the sections HFKM, EFGM intersect each other, therefore $KL\; \backslash cdot\; HL$ has a constant ratio to $LF^2$, consequently (1 cor. 11. Part II.) the section HFKM is an ellipse, of which HK is a diameter and MF an ordinate.

Fig. 78. CASE II. 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 $DI^2\; =\; AI\; \backslash cdot\; IB$,

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.

Fig. 79. CASE III. 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

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.).

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. I. 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. II. A circle. III. An ellipse. IV. A parabola. V. An hyperbola. The two first of these, however, viz. the VOL. VI. Part II.

triangle and circle, may be referred to the hyperbola Of the Curvature of the Conic Sections. 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.

I. A circle is said to touch a conic section in any point, when the circle and conic section have a common tangent in 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.

Let PL be any chord in a circle, PX a tangent at Fig. 80. one of its extremities, and LK a diameter passing through the other extremity: draw any chord Gg parallel to the tangent PX, meeting PL in E, and from its extremities draw GH, g h perpendicular to the diameter, meeting PL in N and n; the square of GE is equal to the rectangle contained by PE and LN, and the square of gE is equal to the rectangle contained by PE and L n.

From G and g draw the straight lines GP, g P, GL, g L, and let LM a perpendicular to the diameter, and therefore a tangent to the circle at L, meet the tangent PX in M. The triangle NGE is evidently similar to the triangle LMP, and $LM\; =\; MP$, therefore $NG\; =\; GE$; hence the angles GNL, GEP are equal. Now the angle PGE is equal to the alternate angle GPX, that is, to the angle GLN in the alternate segment of the circle (32. 3. E.), therefore the triangles PGE, GLN are similar, and

In the same way it may be demonstrated that $n\; g\; =\; g\; E$, and that the triangles PgE, gLn are similar, and therefore that

If a circle be described touching a conic section, and cutting off from the diameter that passes through

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 cuts off from the diameter 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.

Fig. 81, 82, 83, 84. Let $P\backslash rho$ be the diameter of a conic section; let a circle $GPg$ touch the section in $P$ the vertex of the diameter, and cut off from it a segment $PL$, which is either greater or less than the parameter of the diameter; in the former case a part $GPg$ of the circumference of the circle on each side of $P$ the point of contact will be wholly without the conic section, as in fig. 81. and fig. 82. and in the latter a part $GPg$ of the circumference on each side of $P$ will be wholly within the section, as in fig. 83. and fig. 84.

Through $L$ draw $LK$ a diameter of the circle; let $DEd$ an ordinate to the diameter of the section meet the circle in $G$ and $g$, so that the points $G,\; P,\; g$ may be on the same side of $LK$ the diameter of the circle, and draw $GH,\; g\backslash bar\{b\}$, $PO$ perpendicular to $LK$, the two former lines meeting $LP$ in $N$ and $n$. From $L$ towards $P$ place $LR$ in the diameter equal to its parameter; then in the former case the point $R$ will fall between $L$ and $P$, as in fig. 81. and 82.; and in the latter it will fall in $LP$ produced, as in fig. 83. and 84.

CASE I. First, let the section be a parabola (fig. 81. 83.)

Then $DE^2$, also $dE^2\; =\; PE\; \backslash cdot\; RL$. (Cor. prop. 9. of Part I.)
$\backslash left.\; \backslash begin\{array\}\{l\}\; \backslash text\{Now\; \}\; GE^2\; =\; PE\; \backslash cdot\; LN\; \backslash \backslash \; \backslash text\{and\; \}\; gE^2\; =\; PE\; \backslash cdot\; Ln\; \backslash end\{array\}\; \backslash right\backslash \}\; \backslash text\{\; (Lemma).\}$

$\backslash rho\; P\; :\; \backslash rho\; E\; ::\; LR\; :\; LV$,
and therefore $\backslash rho\; P\; :\; LR\; ::\; \backslash rho\; E\; :\; LV\; ::\; \backslash rho\; E\; \backslash cdot\; EP\; :\; LV\; \backslash cdot\; EP$.
But $\backslash rho\; P\; :\; LR\; ::\; \backslash rho\; E\; \backslash cdot\; EP\; :\; DE^2$, or $dE^2$ (12. Part II. and 21. Part III.);
therefore $DE^2$, also $dE^2\; =\; LV\; \backslash cdot\; EP$.
$\backslash left.\; \backslash begin\{array\}\{l\}\; \backslash text\{Now\; \}\; GE^2\; =\; LN\; \backslash cdot\; EP\; \backslash \backslash \; \backslash text\{and\; \}\; gE^2\; =\; Ln\; \backslash cdot\; EP\; \backslash end\{array\}\; \backslash right\backslash \}\; \backslash text\{\; (Lemma).\}$
therefore $DE^2\; :\; GE^2\; ::\; LV\; :\; LN$,
and $dE^2\; :\; gE^2\; ::\; LV\; :\; Ln$.

Now as $P\backslash rho$ and $RL$ are similarly divided at $E$ and $V$, if the point $E$ approach to $P$, the point $V$ will approach to $R$, and as $E$ may come nearer to $P$ than any assignable line, so $V$ may come nearer to $R$ than any assignable line; but as in the same circumstances $GH$ and $g\backslash bar\{b\}$ approach to $PO$, and $N$ and $n$ approach to $P$, it is evident that the ordinate $Dd$ may have such a position that the points $N,\; n$, and the vertex $P$, may be all on the same side of $V$, and the same thing have place for every other position of the ordinate nearer to the tangent; therefore, in these circumstances, when $LP$ the segment cut off from the diameter is greater than $LR$ the parameter (fig. 82.), $LV$ will be less than either $LN$ or $Ln$, and consequently $DE^2$

Therefore $DE^2\; :\; GE^2\; ::\; LR\; :\; LN$,
and $dE^2\; :\; gE^2\; ::\; LR\; :\; Ln$.

Now if the ordinate $Dd$ be supposed to approach to the tangent at the vertex, the points $G,\; g$ will approach to $P$, the lines $GH,\; g\backslash bar\{b\}$ to the line $PO$, and the points $N,\; n$ to the vertex $P$, where they will at last coincide; hence it is evident, that the ordinate $DEd$ may be at such a distance from the tangent that the points $N,\; n$, and the vertex $P$, may be all on the same side of the point $R$; in this position of the ordinate if the segment cut off by the circle be greater than the parameter, as in fig. 81. then $LR$ will be less than either $LN$, or $Ln$, and therefore $DE^2$ less than $GE^2$, also $dE^2$ less than $gE^2$, so that the points $G,\; g$ are both without the parabola. If the ordinate be supposed to approach nearer to the tangent, as the points $N,\; n$ will also approach nearer to $P$, the line $LR$ will still be less than either $LN$, or $Ln$, and therefore $DE^2$ less than $GE^2$, and $dE^2$ less than $gE^2$. Hence it follows, that every point in the arch $GPg$, which lies on each side of the point of contact is without the parabola.

If the segment cut off by the circle be less than the parameter (fig. 83.), and therefore $LR$ greater than either $LN$ or $Ln$, then, reasoning as before, it will appear that $DE^2$ is greater than $GE^2$, and $dE^2$ greater than $gE^2$, so that the points $G,\; g$ are within the parabola; and as the same will hold for every other position of the ordinate nearer to the tangent, the arch $GPg$ which lies on each side of the point of contact is wholly within the parabola.

CASE II. Next, let the section be an ellipse, or an hyperbola (fig. 82. 84.) (A). Take $V$ a point in $LR$, so that

$\backslash rho\; P\; :\; \backslash rho\; E\; ::\; LR\; :\; LV$,
and therefore $\backslash rho\; P\; :\; LR\; ::\; \backslash rho\; E\; :\; LV\; ::\; \backslash rho\; E\; \backslash cdot\; EP\; :\; LV\; \backslash cdot\; EP$.
But $\backslash rho\; P\; :\; LR\; ::\; \backslash rho\; E\; \backslash cdot\; EP\; :\; DE^2$, or $dE^2$ (12. Part II. and 21. Part III.);
therefore $DE^2$, also $dE^2\; =\; LV\; \backslash cdot\; EP$.
$\backslash left.\; \backslash begin\{array\}\{l\}\; \backslash text\{Now\; \}\; GE^2\; =\; LN\; \backslash cdot\; EP\; \backslash \backslash \; \backslash text\{and\; \}\; gE^2\; =\; Ln\; \backslash cdot\; EP\; \backslash end\{array\}\; \backslash right\backslash \}\; \backslash text\{\; (Lemma).\}$
therefore $DE^2\; :\; GE^2\; ::\; LV\; :\; LN$,
and $dE^2\; :\; gE^2\; ::\; LV\; :\; Ln$.

Now as $P\backslash rho$ and $RL$ are similarly divided at $E$ and $V$, if the point $E$ approach to $P$, the point $V$ will approach to $R$, and as $E$ may come nearer to $P$ than any assignable line, so $V$ may come nearer to $R$ than any assignable line; but as in the same circumstances $GH$ and $g\backslash bar\{b\}$ approach to $PO$, and $N$ and $n$ approach to $P$, it is evident that the ordinate $Dd$ may have such a position that the points $N,\; n$, and the vertex $P$, may be all on the same side of $V$, and the same thing have place for every other position of the ordinate nearer to the tangent; therefore, in these circumstances, when $LP$ the segment cut off from the diameter is less than $LR$ the parameter (fig. 84.), $LV$ will be less than either $LN$ or $Ln$, and consequently $DE^2$

less than $GE^2$, also $dE^2$ less than $gE^2$; thus the points $G,\; g$ as well as every other point in the arch $GPg$ which lies on both sides of the vertex are without the ellipse or hyperbola.

On the contrary, when $LP$ is less than $LR$ the parameter (fig. 84.), $LV$ will be greater than either $LN$ or $Ln$, and therefore $DE^2$ greater than $GE^2$, also $dE^2$ greater than $gE^2$; and therefore the points $G,\; g$, as well as every other point in the arch $GPg$, are within the ellipse or hyperbola.

(A) As the reasoning applies alike to the ellipse and hyperbola, to avoid a number of figures, those for the hyperbola are omitted.

Of the Cur- of contact. For if a greater circle be described it will
vature of
the Conic
Sections. cut off from the diameter a segment greater than its
parameter, therefore a part of its circumference on
each side of the point of contact will be wholly with-
out 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 segment less than its parameter,
therefore a part of its 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 curvature.

COR. 2. Only one circle can have the same curva-
ture with a conic section in a given point.

Plate CLXIII. 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 with-
in the conic section; but the circle of curvature
at the vertex of the conjugate axis of an ellipse
falls wholly without the conic section.

Fig. 85, 86, 87, 88. Let $P\backslash rho$ be the axis of a parabola (fig. 85.), and
PGL the circle of curvature at its vertex, which
therefore cuts off from the axis a segment PL equal to
the parameter of the axis; because the tangent at the
vertex is common to the parabola and circle, the cen-
tre of the circle is in $P\backslash rho$. Let $DEd$ an ordinate to
the axis meet the circle in $G$ and $g$; it may be shewn
as in last proposition that

But in every position of the ordinate $LP$ is greater
than $LE$, therefore $DE^2$ is always greater than $GE^2$,
and $dE^2$ greater than $gE^2$; therefore the circle is
wholly within the parabola. Next let $P\backslash rho$ be the
transverse axis of an ellipse or hyperbola (fig. 86, 87.),
or the conjugate axis of an ellipse (fig. 88.), and
PGL the circle of curvature, then as in the para-
bola the centre of the circle will be in the axis. Draw
$Dd$ an ordinate to the axis meeting the circle in $G,\; g$;
and take a point $V$ in $PL$, so that

then it will appear as in last prop. that

Now, when $P\backslash rho$ is the transverse axis of an ellipse,
(fig. 86.) as $P\backslash rho$ is greater than $LP$, and $P\backslash rho\; :\; PL\; ::$
$PE\; :\; PV$, therefore $PE$ is greater than $PV$, and
hence $LV$ is always greater than $LE$, therefore $DE^2$
is greater than $GE^2$, also $dE^2$ greater than $gE^2$, so
that the circle falls wholly within the ellipse.

Again, when $P\backslash rho$ is the transverse axis of an hy-
perbola (fig. 87.), as $\backslash rho\; E$ is greater than $\backslash rho\; P$, there-
fore $LV$ is greater than $LP$, and consequently greater
also than $LE$; hence $DE^2$ is greater than $GE^2$, and
$dE^2$ is greater than $gE^2$, and the circle is wholly
within the hyperbola.

Lastly, When $P\backslash rho$ is the conjugate axis of an el-

lipse (fig. 88.), as $P\backslash rho$ is less than $LP$, and $P\backslash rho\; :\; LP\; ::$
$PE\; :\; PV$, therefore $PE$ is less than $PV$; hence $LV$
is less than $LE$, and consequently $DE^2$ is less than
$GE^2$, also $dE^2$ less than $gE^2$, therefore the circle is
wholly without the ellipse.

The circle of curvature at the vertex of any dia-
meter of a conic section, which is not an axis,
meets the conic section again in one point on-
ly; and between that point and the vertex of
the diameter the circle falls wholly within the
conic section on the one side, and wholly with-
out it on the other.

CASE I. Let the conic section be a parabola, of
which $P\backslash rho$ is a diameter (fig. 89.) and PLK the circle
of curvature at the vertex, cutting off from the dia-
meter a segment PL equal to its parameter. Draw
LK a diameter of the circle, and draw $PO$ perpendi-
cular to $LK$, this line will necessarily meet the circle
again, let it meet the circle in $I$; draw $IS$ parallel to
the tangent at $P$, meeting the chord $PL$ in $S$; then,
because $IP$ is perpendicular to $LK$,

hence (Cor. Prop. 9. Part I.) $I$ is a point in the parabola.

Let $DEd$ an ordinate to the diameter $P\backslash rho$ meet the
arch $PLI$ anywhere in $G$; draw $GH$ perpendicular
to $LK$, meeting $PL$ in $N$, then, because $LP$ is equal
to the parameter, as in Prop. I. Case I.,

But wherever the point $G$ be taken in the arch $PLI$,
$LO$ is greater than $LH$, therefore $DE^2$ is also greater
than $GE^2$; thus the arch $PGLI$ falls wholly within
the parabola.

Let the ordinate $DEd$ now meet the arch $PKI$
anywhere, as at $g$, draw $gb$ perpendicular to $LK$,
meeting $LP$ in $n$, then it will appear as before that

but $LO$ is less than $Lb$, and therefore $dE^2$ less than
$gE^2$, thus the arch $PgKI$ falls wholly without the
parabola.

CASE II. Let the conic section be either an el-
lipse or hyperbola (fig. 90.) of which $P\backslash rho$ is a diam-
eter, and PLK the circle of curvature at its vertex,
cutting off $PL$ equal to its parameter. Draw $LK$ the
diameter of the circle and $LQ$ perpendicular to $LK$,
and let $\backslash rho\; Q$ a tangent to the conic section in $\backslash rho$ meet
$LQ$ in $Q$. Join $PQ$, this line will necessarily meet
the circle again; let it meet the circle in $I$; and
draw $IS,\; IT$ parallel to $QI,\; QI\text{'}$, meeting $PL$ in $S,\; T$.
Because of the parallels,

hence $I$ is a point in the ellipse or hyperbola (13. Prop.
Part II. and 21. Prop. Part III.).

Let $DEd$ an ordinate to the diameter $P\backslash rho$ meet the
arch $PLI$ anywhere in $G$, if the point $L$ is between

Of the Cur-
vatures of
the Conic
Sections.

Let $P$ and $p$, or the arch $PL$, if $L$ is in $Pp$ produced.
Let $Dd$ meet $PI$ in $Y$, draw $GH$ perpendicular to
$LK$ meeting $PL$ in $N$, and $PI$ in $Z$, and draw $YV$
parallel to $GN$ meeting $LP$ in $V$. Because $EY,\; pQ$
are parallel, also $VY,\; LQ$.

now $LP$ being the parameter, we have, as in Case II.
Prop. I.

but wherever the point $G$ be taken in the arch $PGI$,
$QY$ is greater than $QZ$, therefore also $DE^2$ is greater
than $GE^2$; thus the arch $PGI$ falls wholly within the
conic section.

Let the ordinate $DEd$ now meet the other arch
$PgI$ anywhere in $g$; draw $gb$ perpendicular to $LK$
meeting $LP$ in $n$, and $IP$ in $m$, then it will in like
manner appear that

and since in this case $QY$ is less than $Qn$, therefore
$dE^2$ is less than $gE^2$; hence the arch $PgI$ is wholly
without the conic section.

Fig. 91. 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 perpen-
dicular 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 $P$. Be-
cause 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,

Cor. 1. Hence the radius is equal to $\backslash frac\{2FP^2\}\{FK\}$.

Cor. 2. The radius is also equal to $\backslash frac\{2FK^2\}\{AF^2\}$, where
$AF$ is the distance of the focus from the vertex of the
parabola; for $FP\; =\; \backslash frac\{FK^2\}\{AF\}$ (11. Part I.)

Cor. 3. Hence also the radius is equal to
$\backslash frac\{\backslash frac\{1\}\{2\}\; L\; \backslash cdot\; FP^2\}\{FK^2\}$, where $L$ denotes the parameter of the axis,

The radius of the circle of curvature at the vertex
of any diameter of an ellipse, or hyperbola, is Fig. 92.
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
$Qq$ 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, there-
fore,

Secondly, the triangles $PSN,\; PHM$ are similar,
therefore $PN\; :\; PS\; ::\; PM\; :\; PH$;
but $PS\; :\; CQ\; ::\; (CQ\; :\; PO\; ::)\; Qq\; :\; PM$,
therefore $PN\; :\; CQ\; ::\; Qq\; :\; PH$,

or, since $PN=AC$ (18. Part II. and 25. Part III.),

Cor. 1. Hence the radius of curvature is equal to
$\backslash frac\{CQ^2\}\{PS\}$, and the chord passing through the focus is equal

Cor. 2. The radius of curvature is also equal to
$\backslash frac\{CQ^2\}\{AC\; \backslash cdot\; BC\}$ for $PS\; =\; \backslash frac\{AC\; \backslash cdot\; BC\}\{CQ\}$ (14. Part II. and 23. 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
$\backslash frac\{\backslash frac\{1\}\{2\}\; L\; \backslash cdot\; FP^2\}\{FK^2\}$. For the triangles $PFK,\; NPS$ are mani-
festly similar, therefore

This expression for the radius of curvature is the same
for all the three conic sections.

CONICHTHYODONTES,

CONIC SECTIONS.

Plate CLVI.

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

Fig. 5.

Fig. 6.

Fig. 7.

Fig. 8.

Fig. 9.

Fig. 10.

Fig. 11.

Fig. 12.

Fig. 13.

Fig. 14.

Fig. 15.

CONIC SECTIONS.

Plate CLVII.

Fig. 16.

Fig. 17.

Fig. 18.

Fig. 19.

Fig. 20.

Fig. 22.

Fig. 21.

Fig. 23.

Fig. 24.

Fig. 25.

Fig. 26.

Fig. 27.

Fig. 29.

Fig. 28.

CONIC SECTIONS.

Fig. 30.

Fig. 31.

Fig. 32.

Plate CLVIII.

Fig. 33.

Fig. 34.

Fig. 35.

Fig. 36.

Fig. 37.

Fig. 38.

Fig. 39.

Fig. 40.

Fig. 42.

Fig. 43.

Fig. 44.

Fig. 46.

Fig. 45.

Fig. 48.

Fig. 47.

Fig. 49.

Fig. 50.

Fig. 51.

Fig. 53.

Fig. 52.

CONIC SECTIONS.

Plate CLXI.

Fig. 64.

Fig. 70.

Fig. 65.

Fig. 66.

Fig. 67.

Fig. 71.

Fig. 72.

Fig. 68.

Fig. 69.

Fig. 73.

Fig. 75.

Fig. 76.

Fig. 77.

Fig. 78.

Fig. 79.

Fig. 74.

Fig. 81.

Fig. 80.

Fig. 82.

Fig. 83.

Fig. 84 is a geometric diagram showing a circle with a secant line passing through points R, V, N, P, G, L, and P. A vertical line segment connects points D, G, E, and H. A horizontal line segment connects points H, O, L, and P. Other points labeled include A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.

Fig. 85 is a geometric diagram showing a circle with a secant line passing through points D, G, E, L, and P. A vertical line segment connects points D, G, E, and L. A horizontal line segment connects points P, E, L, and P. Other points labeled include A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.

Fig. 86 is a geometric diagram showing a circle with a secant line passing through points D, G, E, L, and P. A vertical line segment connects points D, G, E, and L. A horizontal line segment connects points P, E, L, and P. Other points labeled include A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.

Fig. 88 is a geometric diagram showing a circle with a secant line passing through points D, G, E, L, and P. A vertical line segment connects points D, G, E, and L. A horizontal line segment connects points P, E, L, and P. Other points labeled include A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.

Fig. 87 is a geometric diagram showing a circle with a secant line passing through points D, G, E, L, and P. A vertical line segment connects points D, G, E, and L. A horizontal line segment connects points P, E, L, and P. Other points labeled include A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.

Fig. 89 is a geometric diagram showing a circle with a secant line passing through points D, G, E, L, and P. A vertical line segment connects points D, G, E, and L. A horizontal line segment connects points P, E, L, and P. Other points labeled include A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.

Fig. 91 is a geometric diagram showing a circle with a secant line passing through points D, G, E, L, and P. A vertical line segment connects points D, G, E, and L. A horizontal line segment connects points P, E, L, and P. Other points labeled include A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.

Fig. 92 is a geometric diagram showing a circle with a secant line passing through points D, G, E, L, and P. A vertical line segment connects points D, G, E, and L. A horizontal line segment connects points P, E, L, and P. Other points labeled include A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.

Fig. 90 is a geometric diagram showing a circle with a secant line passing through points D, G, E, L, and P. A vertical line segment connects points D, G, E, and L. A horizontal line segment connects points P, E, L, and P. Other points labeled include A, B, C, D, E, F, G, H, I, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z.