Scholium.

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.

Prop. IV.

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, Pf 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 FP 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 HK k is equal to HkK: now the angles PK k, pkK are right angles, therefore the angle HKP is equal to HkP; but these angles have been shewn to be equal to HFP and HFP respectively, therefore the lines PF and Pf make equal angles with HF.

Prop. V.

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

\[ \frac{HP}{Hp} : \frac{\text{sine } HFP}{\text{sine } HFP}. \]

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 HFP. Now, in the triangle HPQ,

\[ \frac{HP}{HQ} : \frac{\text{sine } HQP}{\text{sine } HQF} : \frac{\text{sine } HFP}{\text{sine } HFP} \] (by Trigon.)

therefore \( \frac{HP}{Hp} : \frac{\text{sine } HFP}{\text{sine } HFP}. \)

Prop. VI.

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 HPh, 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 Hh in L and l, and draw DF, dF, PF to the focus.

Because Hh is parallel to Dd,

\[ \frac{HD}{hd} : \frac{KD}{Kd}. \]

But KD, Kd being tangents to the parabola,

\[ \frac{\text{sine } hDF}{\text{sine } HDF} : \frac{KD}{Kd} = \frac{5}{1}. \]

Therefore, \( \frac{\text{sine } hDF}{\text{sine } HDF} : \frac{HD}{hd} = \frac{5}{1}. \)

Now sine kPF : sine hDF :: hd : hP (5.), Therefore, (23.5.E.) sine hPF : sine HDF :: HD : hP;

but sine HPF, or sine hPF : sine HDF :: HD : hP, therefore the ratio of HD to hP is the same as that of HD to HP, wherefore HP = Ph.

Again, because the angles HDF and hDF are respectively equal to HDL and hdL (3.)

\[ \frac{DH}{dh} : \frac{\text{sine } hDL}{\text{sine } HDL}, \]

Now HL : DH :: sine HDL : sine HLD, or sine hld (by Trigon.)

therefore (23.5.E.) HL : dh :: sine hdl : sine hld :: hl : dh,

wherefore HL, and hd have the same ratio to dl, hence HL = hl; and since it has been shewn that HP = Ph, 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 Hh 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.

Prop. VII.

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 DE'd 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 :: sine HDF : sine HPF (5.) But the angle HDF is equal to HDL, and HPF is equal to HKP (3.), that is (because of the parallel lines DL, PK) to HLD; therefore

PH : HD :: sine HDL : sine 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.

Prop. VIII.

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

Let DE, a straight line passing through the focus, be an ordinate to the diameter PE; the absciss PE is equal to \( \frac{1}{4} \) the parameter, and the ordinate ED 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. 1. E.), and \( PE = PF = \frac{1}{4} \) 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 ED is equal to 4 EP, or to 4 PF, that is, to the parameter of the diameter.

Prop. IX.

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

Let QH, RK, any two diameters, be produced to meet PI, a tangent to the curve at P, in the points G, I; then

\[ HG : KI :: PG^2 : PI^2. \]

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

Now KI = KR, and KL = KM (7.)

Therefore, by subtraction, LI = MR = HQ,

But LO = HN = HG (7.)

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 \( 2IK :: PG : PI \),

Hence, taking the rectangles of the corresponding terms,

\[ 2GH \cdot IO : 2IO \cdot IK :: PG^2 : PI^2 \]

Therefore GH : IK :: PG^2 : PI^2

Cor. The squares of semi-ordinates, and of ordinates to any diameter, are to one another as their corresponding abscisses. Let HE, KN be ordinates to the diameter PN; draw PG a tangent to the curve at the vertex of the diameter, and complete the parallelograms PEHG, PNKI; then PG, PI are equal to EH, NK, and GH, IK to PE, PN, respectively therefore \( HE^2 : KN^2 :: PE : PN \).

Prop. X.

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

Let HB be an ordinate to the diameter PK, the rectangle contained by PB and the parameter of the diameter is equal to the square of HB, the semi-ordinate.

Let DE 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.).

Therefore \( Dd : DE :: DE : PE \),

and \( Dd \cdot PE = DE^2 \),

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

Therefore \( Dd \cdot PB = HB^2 \).

Scholium.

It was on account of the equality of the square of the semi-ordinate to a rectangle contained by the parameter of the diameter and the absciss, that Apollonius called the curve line to which the property belonged a Parabola.

Prop. XI.

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

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

Produce FB and FA to meet the directrix in D and C, and join AB. The lines FC, FD are bisected at A and B (3 cor. 3.), therefore (2. 6. E.) AB is parallel to CD, or perpendicular to CF, and consequently 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 equiangular; hence

\[ FP : FB :: FB : FA. \]

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

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

\[ HP : PK :: PF : FB :: FB : FA :: FD : FC. \] 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 = half the parameter of the axis.

**Prop. XII.**

If an ordinate and absciss 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.

**Fig. 11.**

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 = 2 EG, 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 inscribing 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.

**Prop. XIII. Problem.**

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

**First Method. By Mechanical Description.**

Let AB be the given directrix, and F the focus. Place the edge of the 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.

**Second Method. By finding any number of points in the curve.**

Through the focus F draw EFC perpendicular to the directrix, and EC will be the axis. Draw any straight line HEA 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 HA in D and d; these will be points in the parabola required, as is sufficiently evident.

**Prop. XIV. Problem.**

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

Let DPd be the given parabola; draw any two parallel chords Dd, Ec, and bisect them at H and K; join KH, meeting the parabola in P, the straight line PHK is the diameter (4 cor. 6.) the point P is its vertex, and Dd, Ec, 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.). PART II. OF THE ELLIPSE.

Definitions.

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 AB ab, 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 Absciss.

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

Prop. I.

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

Let AB ab 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 = A a,

Because A, a are the points in the ellipse,

\[ A f + AF = a F + af \] (Def. 1.)

therefore \( F f + 2 AF = F f + 2 af \);

Hence \( 2 AF = 2 af \), and \( AF = af \),

and \( A f + AF = A f \times af = A a \).

But D and A being points in the ellipse,

\[ DF + DF = A f + AF, \]

therefore \( DF + DF = A a \).

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 \( Pf + DF \) is greater than \( DF \) (21. 1. E.), that is, than A a. 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 \( DF \) (21. 1. E.), that is, than A a.

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 B b be the conjugate axis; join \( fb, fb \). Because \( CF = Cf \), and \( Cb \) is common to the triangles \( Cf b, Cf b \), also the angles at C are right angles, these triangles are equal; hence \( fb = fb \), and since \( fb + fb = A a \), \( fb = AC \).

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

Prop. II.

Every diameter of an ellipse is bisected in the centre.

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

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

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

Prop III.

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.

Then \( BC^2 + Cf^2 = Bf^2 = Ca^2 \) (4 cor. 1.).

But because A a is bisected at C,

\[ Ca^2 = Af \cdot fa + Cf^2, \]

therefore \( BC^2 + Cf^2 = Af \cdot fa + Cf^2, \)

and \( BC^2 = Af \cdot fa \).

Prop. Prop. IV.

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 \( HF, HE, HG, fG \); 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. i. E.), so that \( GH + fH = FH + GH \); 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 \( GDE \).

Scholium.

From the property of the ellipse, which forms this last corollary, the points \( F \) and \( f \) take the name of Foci. For writers on optics 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.

Prop. V.

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. i. E.); therefore \( Pf \) is equal and parallel to \( pF \) (33. i. E.), thus \( FP \) is a parallelogram, of which the opposite angles \( P \) and \( p \) are equal (34. i. E.). Now the angles \( FPH, fpk \) are evidently half the supplements of these angles (4 cor. 4.), therefore the angles \( FPH, fpk \), and hence \( CPH, cpk \) 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 \( Fpk, FPk \) 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. i. E.).

Prop. VI.

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 \( fh \); therefore \( PF \) and \( pF \) make equal angles with \( HF \).

Prop. VII.

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 sines of the angles they contain with straight lines drawn from the points of contact to either focus.

Let the straight lines HP, Hp, which intersect each other at H, be tangents to an ellipse at the points Pp,

\[ \text{HP : HQ :: sine HQP or sine HQF : sine HPF (Trigon.)} \]

therefore

\[ \text{HP : Hp :: sine HpF : sine HPF.} \]

Lemma.

Let KLI be a triangle, having its base LI bisected at P; and let Hh, any straight line parallel to the base, and terminated by the sides, be bisected at P; then Pp, 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.

Complete the parallelograms KHPM, KLPN. The triangles KHh, KLI being similar, and Hh, LI similarly divided at P and p,

\[ \text{KH : KL :: Hh : LI :: HP : LP,} \]

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

Next, let Dd meet KP in E, then

\[ \text{HP : DE (: KP : KE) :: Ph : Ed,} \]

therefore DE is equal to Dd.

Prop. VIII.

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 HP, is bisected at E, by Pp, the diameter that passes through the point of contact.

Let Lpl be a tangent at the other extremity of the diameter, and let KD, Kd, tangents at the points D, d, meet the parallel tangents, HP, Lpl in the points H, L, h, l, and draw DF, dF, PF, pF to either focus. Because Hh is parallel to Dd,

\[ \text{HD : hd :: KD : Kd.} \]

But, KD, Kd being tangents to the ellipse,

\[ \text{Sine hdF : sine HDF :: KD : Kd (7),} \]

therefore sine hdF = sine HDF :: HD : hd.

Now, sine hpF : sine hpF : HDF :: HD : HP;

but sine HPF or sine hpF : sine HDF :: HD : HP,

therefore the ratio of HD to hp is the same as that of HD to HP, therefore PH = Ph. In the same manner it may be demonstrated that pL = lp, therefore (Lemma) the diameter pp when produced passes through K, and bisects Dd, which is parallel to Hh or Ll, 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.

Prop. IX.

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 pp, produced in K, and let DEd be an ordinate to that diameter.

Then CE : CP :: CP : CK.

Through P and p, the vertices of the diameter, draw the tangents PH and pL, 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

\[ \text{DH : PH :: sine HPF : sine HDF} \]

and

\[ \text{DL : pL :: sine LpF : sine LDF} \]

Now the angles HPF, LpF are equal (cor. 5.) and the sine of HDF is the same as that of LDF, therefore

\[ \text{DH : PH :: DL : pL,} \]

and by alternation,

\[ \text{DH : DL :: PH : pL;} \]

therefore, because of the parallel lines PH, ED, pL,

\[ \text{EP : Ep :: PK : pK.} \] Take \( CG = CE \), then by division

\[ EG : EP :: Pp : PK, \]

and taking the halves of the antecedents,

\[ CE : EP :: CP : PK; \]

hence, by composition, \( CE : CP :: CP : CK. \)

Cor. 2. The rectangle contained by \( PK \) and \( KP \) is equal to the rectangle contained by \( KE \) and \( KC. \)

For \( KC^2 = PK \cdot KP + CP^2 \) (6. 2. E.)

also \( KC^2 = EK \cdot KC + EC \cdot KC = EK \cdot KC + CP^2 \) (1. 2. E. and by the prop.)

therefore \( PK \cdot KP + CP^2 = EK \cdot KC + CP^2, \)

and \( PK \cdot KP = EK \cdot KC. \)

Prop. X.

If a diameter of an ellipse be parallel to the ordinates to another diameter, the latter diameter shall be parallel to the ordinates to the former.

Let \( PP_p \), a diameter of an ellipse, be parallel to \( DE_d \) any ordinate to the diameter \( Qq \), the diameter \( Qq \) shall be parallel to the ordinates to the diameter \( PP_p \).

Draw the diameter \( dCG \) through one extremity of the ordinate \( dD \), and join \( G \) and \( D \), the other extremity, meeting \( PP_p \) 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 \( PP_p \).

And because \( dG \) and \( dD \) are bisected at \( C \) and \( E \), the diameter \( Qq \) is parallel to \( DG \) (2. 6. E.) therefore \( Qq \) is parallel to any ordinate to the diameter \( PP_p \).

Definitions.

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

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

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

Prop. XI.

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

Let \( DE_d \) be an ordinate to the diameter \( PP_p \), and let \( Qq \) be its conjugate, then

\[ PE \cdot EP : DE_d^2 :: Pp^2 : Qq^2. \]

Let \( KDL \) a tangent at \( D \) meet the diameter in \( K \), and its conjugate in \( L \); draw \( DG \) parallel to \( PP_p \), meeting \( Qq \) in \( G \). Because \( CP \) is a mean proportional between \( CE \) and \( CK \) (9.)

\[ CP^2 : CE^2 :: CK : CE \] (2 cor. 20. 6. E.)

and by division \( CP^2 : PE \cdot EP :: CK : KE. \)

But, because \( ED \) is parallel to \( CL, \)

\[ CK : KE :: CL : DE \text{ or } CG, \]

Cor. 1. The rectangle contained by \( PE \) and \( EP \) is equal to the rectangle contained by \( KE \) and \( CE. \)

For \( PC^2 = KC \cdot CE = KE \cdot EC + EC^2 \) (3. 2. E.)

also \( PC^2 = PE \cdot EP + EC^2 \) (5. 2. E.)

therefore \( KE \cdot EC + EC^2 = PE \cdot EP + EC^2, \)

and \( KE \cdot EC = PE \cdot EP. \)

and because \( CQ \) is a mean proportional between \( CG \) and \( CL \) (9.)

\[ CL : CG :: CQ^2 : CG^2 \text{ or } ED^2; \]

therefore \( CP^2 : PE \cdot EP :: CQ^2 : DE^2, \)

and by inversion and alternation,

\[ PE \cdot EP : DE^2 :: CP^2 : CQ^2 :: Pp^2 : Qq^2. \]

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

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

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

For

\[ DE^2 : de^2 :: AE \cdot Ea : Ae \cdot ea :: HE^2 : he^2, \]

therefore \( DE : de :: HE : he. \)

Cor. 4. If a circle be described upon \( AA \) the transverse axis as a diameter, and \( DE \), any ordinate to the axis, be produced to meet the circle in \( H \), \( HE \) shall be to \( DE \) as the transverse axis \( AA \) to the conjugate axis \( BB. \) For, produce the conjugate axis to meet the circle in \( K \), then, by last corollary,

\[ HE : DE :: KC, \text{ or } AC : BC :: AA : BB. \]

Cor. 5. And if \( HE \) be divided at \( D \), so that \( HE \) is to \( DE \), as the transverse axis to the conjugate axis, \( D \) is a point in the ellipse, and \( DE \) a semi-ordinate to the axis \( AA. \)

Prop. XII.

The transverse axis of an ellipse is the greatest of all its diameters, and the conjugate axis is the least of all its diameters.

Let \( AA \) be the transverse axis, \( BB \) the conjugate axis, and \( CD \) any semidiameter. Draw \( DE \) perpendicular to \( AA \), and \( DL \) perpendicular to \( BB. \)

Because \( AA^2 : BB^2 :: AE \cdot Ea : DE^2 \) (11.)

and \( AA^2 \) is greater than \( BB^2, \)

therefore \( AE \cdot Ea \) is greater than \( DE^2; \)

and \( AE \cdot Ea + EC^2 \) is greater than \( DE^2 + EC^2, \)

that is, \( AC^2 \) is greater than \( DC^2, \)

therefore, \( AC \) is greater than \( DC. \) By the same manner of reasoning it may be shewn that because $Bb^2$ is less than $Aa^2$,

$$BL \cdot Lb + CL^2 \text{ is less than } DL^2 + CL^2;$$

that is, $BC^2$ is less than $DC^2$, and $BC$ less than $DC$.

**Prop. XIII.**

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

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

$$PP : Qq :: Qq : PG,$$

therefore $PP : PG :: PE^2 : Qq^2$ (2 cor. 20. 6. E.)

But $PP^2 : Qq^2 :: PE : DE^2$, (11.)

therefore $PE \cdot EP : DE^2 :: PP : PG.$

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

For $PE \cdot EP : DE^2 :: PP : PG,$

and $PP : PG :: PE : PE \cdot EP : PE \cdot EM,$

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

**Scholium.**

If the rectangles $PGLP$, $HGKM$ 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.

$$CE^2 + CG^2 = CR^2,$$

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. i. E.), $CP^2 + CQ^2 = CR^2 + CS^2$,

therefore $PP^2 + QQ^2 = RR^2 + SS^2$.

**Prop. XIV.**

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

Let $PP$, $QQ$ be two conjugate diameters, and $PE$, Fig. 28. $QG$ semi-ordinates to any third diameter $RR$, then $CG^2 = RE \cdot ER$, and $CE^2 = RG \cdot GR$.

Draw the tangents $PH$, $QK$ meeting $RR$ in $H$ and $K$. The rectangles $HC \cdot CE$ and $KC \cdot CG$ are equal, for each is equal $CR^2$ (9.), therefore

$$HC : CK :: CG : CE,$$

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

$$HC : CK :: HE : CG,$$

wherefore $CG : CE :: HE : CG$,

consequently $CG^2 = CE \cdot EH$ (1 cor. 9.) $RE \cdot ER$.

In like manner it may be shewn that $CE^2 = RG \cdot GR$.

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

For $RR^2 : SS^2 :: RE \cdot ER$, or $CG^2 : PE^2$,

therefore $RR : SS :: CG : PE$.

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.

For $CE^2 + CG^2 = CR^2 + RE \cdot EG = CR^2$.

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

Let $RR$, $SS$ be the axes, and $PP$, $QQ$ any two conjugate diameters; draw $PE$, $QG$ perpendicular to $RR$, and $PL$, $QM$ perpendicular to $SS$. Then

Because $CK : CA :: CA : CL$ (9.)

and $CA : CB :: CL : QM$ (1 cor. 14.)

ex aequo, $CK : CB :: CA : QM$,

therefore $CK \cdot QM = CB \cdot CA$.

But $C \cdot QM =$ twice trian. $CKQ =$ paral. $CPEQ$, therefore the parallelogram $CPEQ = CB \cdot CA$,

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

**Prop. XVI.**

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 points of contact and the parallel tangents, is equal to the square of the semi-diameter to which it is parallel.

Let PH, p h tangents at the vertices of a diameter Pp meet HD h, a tangent to a curve at any point D, in H and h; let CQ be the semi-diameter to which the tangents PH, p h are parallel, and CR that to which H h is parallel, then,

\[ PH \cdot p h = CQ^2, \text{ and } DH \cdot Dh = CR^2. \]

If the tangent HD h 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.

Because \( LP \cdot Lp = LE \cdot LC \) (2 cor. 9.), \( LP : LE :: LC : Lp, \)

hence, and because of the parallels PH, ED, CK, p h,

\[ PH : ED :: CK : p h, \] wherefore \( PH \cdot p h = ED \cdot CK, \) but \( ED \cdot CK = CG \cdot CK = CQ^2 \) (9.) therefore \( PH \cdot p h = CQ^2. \)

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

therefore \( PE : HD :: (LE : LD :: ) CM : CR, \) also \( p E : h D :: (LE : LD :: ) CM : CR, \) hence, taking the rectangles of the corresponding terms,

\[ PE \cdot p E : HD \cdot h D :: CM^2 : CR^2. \]

But if CD be joined, the points D and R are evidently the vertices of two conjugate diameters (cor. Def. 12.) and therefore \( PE \cdot p E = CM^2 \) (14.)

therefore \( HD \cdot h D = CR^2. \)

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, p h meet LK in H and h, and draw DE a semi-ordinate to Pp. Because of the parallels PH, ED, CK, p h,

\[ LE : LD :: EP : DH, \] and \( EC : DK :: Ep : Dh, \) therefore \( LE \cdot EC : LD \cdot DK :: EP \cdot Ep : DH \cdot Dh. \)

But \( LE \cdot EC = EP \cdot Ep \) (1 cor. 9.) therefore \( LD \cdot DK = DH \cdot Dh = \) (by this prop.) \( CR^2. \)

Prop. XVII.

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 DPd be a tangent to the curve at P, and FD, f'd 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/f'P, and produce FD, f'D till they intersect in F. The triangles FDP, EDP, have the angles at D right angles, and the angles FPD, 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 F/Ef', FCD are similar,

therefore \( Ef : fEC \text{ or } Aa :: FC : CD, \) but FC is half of Ff', therefore CD is half of Aa. In like manner it may be shewn that C d is half of Aa.

Cor. If the diameter Qg be drawn parallel to the tangent Dd, it will cut off from PF, P'f' the segments PG, Pg, each equal to AC half the transverse axis. For C dPG, CD Pg are parallelograms, therefore \( PG = dC = AC, \text{ and } Pg = DC = AC. \)

Prof. XVIII.

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 DPd be a tangent, and FD, f'd perpendiculars from the foci, the rectangle contained by FD and f'd is equal to the square of CB half the conjugate axis.

It is evident from the last proposition that the points D, d are in the circumference of a circle whose centre is the centre of the ellipse, and radius GA, half the transverse axis; now F'Dd 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 \( FC = f'C, \text{ and } CH = C d, \text{ and the angles } FCH, f'C d \text{ are equal, } FH \text{ is equal to } f'd, \text{ therefore } \)

\[ DF \cdot df = DF \cdot FH = AF \cdot Fa \] (35. 3. E.) \( = CB^2 \) (3.)

Cor. If FP, P'f be drawn from the point of contact to the foci, the square of FD is a fourth proportional to \( fP, FP \text{ and } BC^2. \) For the lines \( fP, FP \text{ make equal angles with the tangent } \) (4 cor. 4,) and \( f'dP, FDP \text{ are right angles, therefore the angles } f'Pd, FPD \text{ are similar, and } \)

\[ fP : FP :: f'd : FD : f'd : FD \text{ or } CB^2 : FD^2. \]

Prof. XIX.

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

Produce the axes to meet the tangent in M and m, and from D draw the semi-ordinates DE, Dc, which will be perpendicular to the axes. The triangles DEH, CLm are evidently equi-angular, therefore

\[ DH : DE :: Cm : CL, \] hence \( CL \cdot DH = DE \cdot Cm; \) but \( DE \cdot Cm, \) or \( Cc \cdot Cm = BC^2, \) therefore \( CL \cdot DH = BC^2. \)

In the same way it is shewn that \( CL \cdot D h = AC^2. \)

Cor. 1. If a perpendicular be drawn to a tangent at the point of contact, the segments intercepted between the points of contact, and the axes, are to each other

\[ GD : DN :: HD : DK, \] and hence \( GD \cdot DK = HD \cdot DN. \)

But \( GD = AC \) (cor. 17.) and \( ND = CL, \) therefore \( AC \cdot DK = HD \cdot CL = (by the prop.) CB^2, \) wherefore \( AC : BC :: BC : DK, \) hence \( DK \) is half the parameter of \( Aa \) (Def. 13.).

Definitions.

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 GH, 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.

Prop. XX.

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.

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 = Df, then CL is evidently half the difference between AL and aL, or FD and fD, and CE half the difference between fE and FE, and because

\[ Df + DF : fE :: FE - FE : Df - DF \] (Trigon.)

By taking the halves of the terms of the proportion

\[ CA : CF :: CE : CI, \] But \( CA : CF :: CG : CA \) (Def. 14.) therefore \( CG : CA :: CE : CI, \) hence (20. 5. E.) \( EG : AL :: CG : CA :: CA : CF, \) that is, \( DK : DF :: CA : CF. \)

Cor. 1. If the tangent GMN be drawn through M, the extremity of the ordinate passing through the focus, and 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.

\[ FM : FD :: MO : DK :: GF : GE :: FM : EN, \] therefore \( ED = EN. \)

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

For \( AC^2 : BC :: CL \cdot Dh : CL \cdot DH :: Dh : DH. \)

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

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

\[ GD : DN :: HD : DK, \] and hence \( GD \cdot DK = HD \cdot DN. \)

But \( GD = AC \) (cor. 17.) and \( ND = CL, \) therefore \( AC \cdot DK = HD \cdot CL = (by the prop.) CB^2, \) wherefore \( AC : BC :: BC : DK, \) hence \( DK \) is half the parameter of \( Aa \) (Def. 13.).

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.

Prop. XXI.

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-axis 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,

\[ GE : DE :: CG : HD :: CA : CB, \] therefore \( DE \) is a semi-ordinate, and \( D \) a point in the ellipse (5 cor. 11.)

Scholium.

The instrument called the trammels, also the elliptic compasses, 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.

Prop. XXII.

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. 35. is also the diameter of the circle. Draw DE, D'E', D''E'' any number of perpendiculars to the axis, meeting the ellipse in D, D', D'', and the circle in d, d', d'', and join AD, DD', D'D'', D''a; also A d, d d', Of the \(d'd'\), \(d''a\), and draw DG, \(dg\) parallel to Aa, meeting Hyperbola, \(d'f'\) in G and g.

The triangle A d E is to the triangle ADE as d E to DF, that is (4 cor. II.) as the transverse axis to the conjugate axis. Again, because \(d'E'\) and \(d'E\) are similarly divided at D' and D (3 cor. II.)

\[dE : DE :: (d'E' - dE : DE') - DE :: dg : DG,\]

But, triangle \(dg'd'\) : triangle \(DGD'\) :: \(dg : DG\),

therefore the triangles \(dg'd'\), \(DGD'\) as well as the rectangles \(d'E'\), \(DE'\) are to each other as \(d'E\) to DE, or as the transverse axis to the conjugate axis, and consequently the trapezoids \(dEE'd'\), \(DE'D'\) are to each other in the same ratio. In like manner it may be shown, that the trapezoids \(d'E'E'a\), \(DE'E'a\) also the triangles \(d'E'a\), \(DE'E'a\) are to each other as the transverse to the conjugate axis, and therefore the whole rectilineal figure \(A'd'd'a\) inscribed in the semicircle to the whole figure \(ADD'D'a\) 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 A a remaining common to both figures, if we suppose the number of perpendiculars \(dDE\), \(d'DE'\), &c., indefinitely increased, it is evident that the polygons \(A'd'd'a\), \(ADD'D'a\) 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.**

Let A a b 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. I.).

**Part III. Of the Hyperbola.**

**Definitions.**

I. If two points F, f be given in a plane, and a point D be conceived to move in such a manner that \(DF = DF'\), the difference of its distances from them is always the same, the point D will describe upon the plane a line DAD' called an Hyperbola. By assuming first one of the given points F, and then the other f as that to which the moving point is nearest, the difference of the lines DF and DF' in both cases being the same, there will be two hyperbolas DAD', da d', described; 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... the 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 b 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.

**Prop. I.**

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

Let DAD', \( da \) 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,

\[ Af - AF = aF - af \quad (\text{Def. 1.}) \]

therefore \( Ff - 2AF = Ff - 2af \);

Hence \( 2AF = 2af \) and \( AF = af \),

and \( Af - AF = Aa \).

But D and A being points in the hyperbola,

\( Df - DF = Af - AF \), therefore \( Df - DF = Aa \).

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

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

**Lemma I.**

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 \( AB - BC \).

**Prop. II.**

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

Let \( Pp \) be a transverse diameter, it is bisected in 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 \( Ep, EQ \).

Because \( fC = CF \), and CD is parallel to \( EF, fD = DF \) (2. 6. F.). 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 \( FG = Cf, PC = CQ \), and the angles at C equal, are in all respects equal, therefore \( FP = fQ \). In like manner it appears that \( Pf = QF \), therefore \( FQ - fQ \) is equal to \( FP - FP \), or (Def. 1.) to \( FP - fP \); that is, \( FQ - QF \) 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.

**Prop. III.**

The square of half the conjugate axis of an hyperbola is equal to the rectangle contained by the straight lines between either focus and the extremities of the transverse axis. Draw a straight line from \(a\), either of the extremities of the transverse axis, to \(B\), either of the extremities of the conjugate axis.

Then \(BC^2 + Ca^2 = Ba^2 = C^2\) (Def. 7.)

But because \(Aa\) is bisected at \(C\), and produced to \(f\),

\[C^2 = Af/f + Ca^2 \quad (6. 2. E.)\]

therefore \(BC^2 + Ca^2 = Af/f + Ca^2\),

and \(BC^2 = Af/f.\)

**Prop. IV.**

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

Let \(D\) be 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 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 \(FDf\),

\[FD : fD :: FE : fE \quad (3. 6. E.)\]

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.

---

**Scholium.**

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

**Prop. V.**

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

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

**Cor. 1.** If tangents be drawn to an hyperbola at the vertices of a transverse diameter, straight lines drawn from either focus to the points of contact make equal angles with these tangents. For the angle \(Fph\) is equal to \(FPH\).

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

**Prop. VI.**

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 \(HP, Hp\) be tangents to an hyperbola 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 \(PF, fp\) make equal angles with \(HF\).

Draw \(Pf, pf\) to the other focus. In \(PF\) and \(pf\) take \(PK = Pf\), and \(pk = pf\); join \(HK, Hk\), and let \(fK, fk\) 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 shown, that \(fH\) is equal to \(kh\), therefore \(HK\) is equal to \(Kh\); now \(FK\) is equal to \(fK\); for each is equal to the difference between \(FP\) and \(fP\), or \(FP\) and \(fp\), that is, to the transverse axis; therefore the triangles \(FKH, fKh\) are in all respects equal, and hence the angle \(KFH\) is equal to \(fKH\), therefore \(PF\) and \(pf\) 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.

Let HP, Hp, which intersect each other at H, be tangents to an hyperbola, or opposite hyperbolas, at the points P, p; and let PF, pF be drawn to either focus,

\[ \text{HP} : \text{Hp} :: \text{sine} \, \text{HPF} : \text{sine} \, \text{PHF}. \]

Join HF, and in FP take FQ equal to FP, and join HQ; then, the angles at F being equal (6.), the triangles HFQ, HFp are equal, therefore HQ is equal to Hp, and the angle HQF is equal to HPF. Now in the triangle HQP,

\[ \text{HP} : \text{HQ} :: \text{sine} \, \text{HQP} : \text{sine} \, \text{HPF} : \text{sine} \, \text{HPF} \quad (\text{Trig.}) \]

therefore

\[ \text{HP} : \text{Hp} :: \text{sine} \, \text{HPF} : \text{sine} \, \text{PHF}. \]

Lemma II.

Let KL l be a triangle, having its base L l bisected at p, and let H h, any straight line parallel to the base, and terminated by the sides produced, be bisected at P, then P, p the point of bisection, and K the vertex of the triangle, are in the same straight line, and that line bisects D d any other line parallel to the base.

Join KP, Kp. The triangles KH h, KL l being similar, and H h, L l similarly divided at P, p,

\[ \text{KH} : \text{KL} :: (\text{Hh} : \text{Ll} ::) \text{HP} : \text{Lp}. \]

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

Next let D d meet Kp in E, then

\[ \text{HP} : \text{DE} :: \text{PK} : \text{EK} :: \text{Ph} : \text{Ed}, \]

therefore DE is equal to Ed.

Prop. VIII.

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.

The straight line D d, terminated by the hyperbola, and parallel to the tangent HP h, is bisected at E by PP the transverse diameter produced, which passes through P, the point of contact.

Let L p 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 HP h, Lp l in the point H, L, h, l, and draw DF, dF, PF to either focus. Because H h is parallel to D d,

\[ \text{HD} : \text{hd} :: \text{KD} : \text{Kd}. \]

But KD, Kd being tangents to the hyperbola,

\[ \text{sine} \, \text{hdF} : \text{sine} \, \text{HDF} :: \text{KD} : \text{Kd} \quad (7.) \]

therefore sine hdF : sine HDF :: HD : hd,

now, sine hpF : sine hdF :: hd : hp \quad (7.)

therefore, (23. 5. E.) sine hpF : sine HDF :: HD : AP;

but sine HPF or sine hpF : sine HDF :: HD : HP,

therefore the ratio of HD to hp is the same as the ratio of HD to HP, wherefore PH = Ph. In the same manner it may be demonstrated that pL = lp, therefore (lemma 2.) the diameter Pp when produced passes through K, and bisects D d, which is parallel to H h, or L l, 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.

Prop. IX.

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 Pp in K, and let DE d be an ordinate to that diameter,

Then CE : CP :: CP : CK.

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

\[ \text{DH} : \text{HP} :: \text{sine} \, \text{HPF} : \text{sine} \, \text{HDF}, \]

and DL : Lp :: sine LpF : sine LDF, or sine HDF \quad (7.)

Now the angles HPF, LpF are equal (1 cor. 5.); therefore,

DH : PH :: DL : pL,

and by alternation

DH : DL :: PH : pL;

therefore, because of the parallel lines PH, ED, pL,

EP : Ep :: PK : pK.

Take CG = CE, then PG = Ep, and by composition

EG : EP :: Pp : PK,

and taking the halves of the antecedents

For KC² = CP² - PK · Kp (5. 2. E.)

also KC² = EC · KC - EK · KC = CP² - EK · KC (3. 2. E. and by the prop.)

therefore CP² - PK · Kp = CP² - EK · KC,

and PK · Kp = EK · KC.

PROP. X.

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

Let DH, a tangent to the hyperbola at D, meet the conjugate axis Bb in H, and let DG be perpendicular to that axis, then

CG : CB :: CB : CH.

Let DH meet the transverse axis in K, draw DE perpendicular to that axis, draw DF, Df to the foci, and describe a circle about the triangle DfF; the conjugate axis will evidently pass through the centre of the circle, and because the angle 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 LDH, therefore,

LG : GD (=CE) :: CK : CH;

hence LG · CH = CE · CK = (by last prop.) CA².

Now LC · CH = CF² (3. 3. E.)

therefore LC · CH = LG · CH = CF² - CA²,

that is, CG · CH = CB² (Def. 7.)

wherefore CG : CB :: CB : CH.

DEFINITION.

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

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

CE : EP :: CP : PK;

hence, by division, CE : CP :: CP : CK.

Cor. 1. The rectangle contained by PE and Ep is equal to the rectangle contained by KE and CE.

For CP² = KC · CE = EC² = KE · EC (2. 2. E.)

also CP² = EC² = PE · Ep (6. 2. E.)

therefore EC² = KE · EC = EC² = PE · Ep,

and KE · EC = PE · Ep.

Cor. 2. The rectangle contained by PK and Kp is equal to the rectangle contained by KE and KC.

A h, or to BC; also aH' is equal to AH, or to BC; hence, by the definition, CH' and C'H' are asymptotes of the opposite hyperbola d a 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' being equal and parallel, the points H, B, H' are in a straight line passing through B parallel to A a; the same is true of the points h, b, H'.

PROP. XI.

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 · 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

CK : CE :: CA² : CE² (2 cor. 20. 6. E.)

Bot CK : CE :: LC : LG,

and CA² : CE² :: AH² : EM²;

therefore LC : LG :: AH² : EM².

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

LC : CG :: CB² : CG²,

and therefore

LC : LG :: CB² : CB² + CG², or CB² + ED²;

wherfore AH² : EM² :: CB² : CB² + ED²;

Now AH² = CB² (Def. 11.)

therefore EM² = CB² + ED²,

consequently EM² is greater than ED², and EM greater The greater than E.D., 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,

\[ ME' = MD \cdot DM + DE^2; \]

but it has been shewn that

\[ ME' = BC^2 + DE^2, \]

therefore \( MD \cdot DM = BC^2. \)

Cor. 1. Hence, if in a straight line Mm, terminated by the asymptotes, and parallel to the conjugate axis, there be taken a point D such that the rectangle \( MD \cdot DM \) is equal to the square of that axis, the point D is in the hyperbola.

Cor. 2. If straight lines MDm, NRn, be drawn through D and R, any points in the hyperbola, or opposite hyperbolas, parallel to the conjugate axis, and meeting the asymptotes in M, m, and N, n, the rectangles \( MD \cdot DM, NR \cdot RN \) are equal.

Prop. XII.

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

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

Because NO² is greater than ME², and NR² = MD · DM, (2 cor. ii.)

therefore NO² - NR² = ME² - MD · DM,

that is, RO² is greater than DE²,

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. ii.)

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 Ah, 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.

Scholium.

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

Prop. XIII.

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 opposite hyperbolas, let parallel lines EDc, HGh be drawn to meet the asymptotes in E, e, and H, h, the rectangles ED · De, HG · Gh are equal.

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

therefore DL : DE :: GM : GH,

and De : De :: Gm : Gh;

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

\[ LD \cdot Di : ED \cdot De :: MG \cdot Gm : HG \cdot Gh. \]

But LD : Di = MG : Gm (2 cor. ii.)

therefore ED · De = HG · Gh.

Cor. 1. If a straight line be drawn through D, d, fig. 54, 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 · De, HG · Gh have been proved to be equal, it may be shewn that the rectangles Ed · de, HG · Gh are equal, therefore ED · De = Ed · de.

Let Ec be bisected in O, then ED · De = EO² - OD²,

and Ed · de = EO² - Od², therefore EO² - OD² = EO² - Od²; 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 Ec will thus become a tangent to the curve at P. Therefore any tangent KPk, which is terminated by the asymptotes, is bisected at P, the point of contact.

Cor. 3. And if any straight line KPk, limited by the asymptotes, be bisected at P a point in the curve, that line is 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 KPk, and terminated by the asymptotes at E and e, the rectangle ED · 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 · De is equal to the square of half the diameter. The demonstration is the same as in the proposition. CONIC SECTIONS.

PROP. XIV.

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 two first lines will be equal to the rectangle contained by the other two lines.

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

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

\[ ED : DH :: Ed : dh, \] and \( eD : DK :: ed : dk, \)

therefore taking the rectangles of corresponding terms,

\[ ED \cdot De : HD \cdot DK :: Ed \cdot de : hd \cdot dk; \]

but \( ED \cdot De = Ed \cdot de (13), \) therefore \( HD \cdot DK = hd \cdot dk. \)

Cor. 1. If the lines DK', D'H', d'k', d'k', be parallel to the asymptotes, and thus form the parallelograms DK'C'H', d'k'C'H', these are equal to one another (16. and 14. 6. E.). And if D'C, d'C be joined, the halves of the parallelograms, or the triangles D'K'C, d'k'C are also equal.

Cor. 2. If from D', d', any two points in an hyperbola, straight lines D'K, d'k be drawn parallel to one asymptote, meeting the other in K' and k', these lines are to each other reciprocally as their distances from the centre, or \( D'K : d'k' :: Ck' : CK'. \) This appears from last cor. and 14. 6. E.

Definitions.

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

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

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

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

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

PROP. XV.

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

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

Let Dd meet the asymptotes in G and g, and let the tangent meet them in K and k. The straight lines Gg, Kk are evidently similarly divided at E and Q, and since KQ = Qk (2 cor. 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. 8.

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.

PROP. XVI.

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

Let Pp, a transverse diameter of an hyperbola, be parallel to DEd, any ordinate to the second diameter Qq, the second diameter Qq 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 Qq is parallel to DG (2.6. E.), therefore Qq is parallel to any ordinate to the diameter Pp.

Definitions.

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

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

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

PROP. XVII.

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 hyperbola, HPa a tangent at its vertex, meeting the asymptotes... the totes in H and h, and Q q the diameter which is conjugate to P p; the tangent H h is equal to the diameter Q q.

Through D, any point in the hyperbola, draw a straight line parallel to the tangent and diameter, cutting either of the conjugate hyperbolae in d, and through D and d draw lines parallel to B b the conjugate axis, meeting the asymptotes in the points K, k, and L, l. The triangles DEK, d EL are similar, as also D k, c d l, therefore

\[ KD : DE :: Ld : dE, \] and \( kD : Dc :: ld : de; \)

therefore, taking the rectangles of the corresponding terms,

\[ KD \cdot Dk : ED \cdot De :: Ld \cdot dl : Ed \cdot de. \]

But \( KD \cdot Dk = BC^2 \) (11.) and \( BC = Ld \cdot dl \) (5 cor. 13.)

therefore \( ED \cdot De = HP^2 \) (4 cor. 13.)

Now \( ED \cdot De = QC^2 \) (5 cor. 13.)

therefore \( HP^2 = QC^2, \) and \( HP = QC, \) and consequently \( H h = Q q. \)

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

\[ CE^2 : CP^2 :: EM^2 : PH^2, \] and \( CE, \) or \( DG : CK :: LG : LC; \) therefore \( EM^2 : PH^2 :: LG : LC, \) and by division, &c. \( EM^2 - PH^2 : PH^2 :: CG : LC :: CG^2 : CG \cdot LC. \)

But since \( PH^2 = MD \cdot DM \) (4 cor. 13.), \( EM^2 - PH^2 = ED^2 = CG^2, \) therefore \( PH = CG \cdot LC; \) wherefore, and since \( PH = CQ \) (17.)

\[ CG : CQ :: CQ : CL. \]

CL : CG :: CQ^2 : CG^2, or \( DE^2, \) therefore \( CP^2 : PE \cdot EP :: CQ^2 : DE^2, \) and by inversion and alternation,

\[ PE \cdot EP : DE^2 :: CP^2 : CQ^2 :: PP^2 : Qq^2. \]

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 Q q. It has been shewn that

\[ CG^2 : CQ^2 :: PE \cdot EP : CP^2, \] therefore, by comp.

\[ CQ^2 + CG^2 : CQ^2 :: CE^2 \text{ or } DG^2 : CP^2, \] and by alter.

\[ CQ^2 + CG^2 : CE^2 :: CQ^2 : CP^2 :: Qq^2 : PP^2. \]

Cor. 2. The squares of semi-ordinates, and of ordinates to any transverse diameter, are to one another as the rectangles contained by the corresponding abscisses; and the squares of semi-ordinates, and of ordinates, Of the nates to any second diameter are to one another as the Hyperbola 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.

PROP. XX.

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.

Fig. 64. Let Rr be the transverse axis, Pp any other transverse diameter, draw PE perpendicular to Rr; then CE being greater than CR, and CP greater than CE, much more is CP greater than CR, therefore Pp is greater than Rr. In like manner it is shewn that if Ss be the conjugate axis, and Qq any other second diameter, Qq is greater than Ss.

PROP. XXI.

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

Fig. 63. Let DE be a semi-ordinate to the transverse diameter Pp; let PG be the parameter of the diameter, and Qq the conjugate diameter. By the definition of the parameter (Def. 16.)

\[ \frac{Pp}{Qg} = \frac{Qg}{PG}, \]

therefore, \( Pp : PG :: Pp^2 : Qg^2 \) (2 cor. 20. 6. E.)

But \( Pp^2 : Qg^2 :: PE : EP : DE^2 \) (19.) therefore \( PE \cdot EP : DE^2 :: Pp \cdot PG \).

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

For \( PE \cdot EP : DE^2 :: Pp : PG, \)

and \( Pp : PG :: EP : EM :: PE \cdot EP : PE \cdot EM, \)

therefore \( DE^2 = PE \cdot EM. \)

Scholium.

If the rectangles PGLp, HGKM be completed, it will appear that the square of ED is equal to the rectangle MP, which rectangle is greater than the rectangle KP, contained by the abscissa 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 abscissa and parameter that Apollonius gave the curve to which the property belonged the name of Hyperbola.

PROP. XXII.

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

Draw the tangents PH, QK, meeting Rr in H and K. The rectangles HC \cdot CE and KC \cdot CG are equal, for each is equal to \( CR^2 \) (9.) therefore,

\[ HC : CK :: CG : CE. \]

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 \( CG^2 = CE \cdot EH = (1 cor. 9.) BE \cdot ER. \)

Again, from the similar triangles HPC, CQK,

\[ HC : CK :: CE : KG. \]

Now it was shewn that \( HC : CK :: CG : CE, \)

therefore \( CG : CE :: CE : KG, \)

consequently

\[ CE^2 = CG \cdot GK = (3. 2. E.) CG^2 + GC \cdot CK. \]

But \( GC \cdot CK = CR^2 \) (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 \( Rr^2 : Ss^2 :: RE \cdot ER, \) or \( CG^2 : PE^2, \)

therefore \( Rr : Ss :: CE : QG. \)

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 \( CE^2 = CG^2 + CR^2; \)

therefore \( CE^2 - CG^2 = CR^2. \)

Cor. 3. The difference of the squares of any two conjugate diameters is equal to the difference of the squares of the axes. Let Rr, Ss be the axes, and Pp, Qq any two conjugate diameters; draw PE, QG perpendicular to Rr, and PL, QM perpendicular to Ss. Then

\[ CE^2 - CG^2 = CR^2, \]

and \( CM^2 = CL^2, \) or \( GB^2 = PE^2 = CS^2, \)

therefore \( CE^2 + PE^2 = (CG^2 + QG^2) = CR^2 - CS^2, \)

that is (47. 1. E.) \( CP^2 = CQ^2 = CR^2 - CS^2, \)

therefore \( Pp^2 - Qq^2 = Rr^2 - Ss^2. \) Prop. XXIII.

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 \( P, p, Q, q \) be any two conjugate diameters, a parallelogram \( DEGH \) formed by tangents to the conjugate hyperbolas at their vertices is equal to the rectangle contained by \( Aa, Bb \) the two axes.

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

Because \( CK : CA :: CA : CL \) (9.) and \( CA : CB :: CL : QM \) (1 cor. 22.) ex aeq. \( CK : CB :: CA : QM \), therefore \( CK \cdot QM = CB \cdot CA \).

But \( CK \cdot QM = \text{twice trian. } CKQ = \text{paral. } CPEQ \), therefore the parallelogram \( CPEQ = CB \cdot CA \);

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

Prop. XXIV.

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 \( PH, p, h \), tangents at the vertices of a transverse diameter \( Pp \), meet \( DH, h \), a tangent to the curve at any point \( D \), in \( H \) and \( h \); let \( CQ \) be the semi-diameter to which the tangents \( PH, p, h \) are parallel, and \( CR \) that to which \( H, h \) is parallel; then

\[ PH \cdot p = CQ^2, \quad \text{and} \quad DH \cdot h = CR^2. \]

Let \( H, h \) meet the semi-diameters \( CP, CQ \) in \( L \) and \( K \). Draw \( ED, RM \) parallel to \( CQ \), and \( DG \) parallel to \( CP \).

Because \( LP : LE :: LC : LP \); \[ LP : LE :: LC : LP; \]

Hence, and because of the parallels \( PH, ED, CK, p, h \),

\[ PH : ED :: CK : p, h, \] wherefore \( PH \cdot p = ED \cdot CK. \)

But \( ED \cdot CK = CG \cdot CK = CQ^2 \) (18.) therefore \( PH \cdot p = CQ^2. \)

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

therefore \( PE : HD :: (LE : LD :: ) CM : CR, \) also \( pE : hD :: (LE : LD :: ) CM : CR, \) hence taking the rectangles of the corresponding terms,

\[ PE \cdot pE : HD \cdot hD :: CM^2 : CR^2. \]

But, if \( CD \) be joined, the points \( D \) and \( R \) are evidently the vertices of two conjugate diameters (cor. Hyperbola def. 15.) and therefore \( PE \cdot pE = CM^2 \) (22.)

therefore \( HD \cdot hD = CR^2. \)

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 \( PH, p, h \) meet \( LK \) in \( H \) and \( h \), and draw \( DE \) a semi-ordinate to \( Pp \). Because of the parallels \( ED, PH, CK, p, h \),

\[ LE : LD :: EP : DH, \] and \( EC : DK :: E_p : Dh, \) therefore

\[ LE \cdot EC : LD \cdot DK :: EP \cdot E_p : DH \cdot Dh. \]

But \( LE \cdot EC = EP \cdot E_p \) (1 cor. 9.) therefore \( LD \cdot DK = DH \cdot Dh = \text{(by this prop.) } CR^2. \)

Prop. XXV.

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 \( ED, Fig. 67. \) \( f'd \) 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 \( FPD, EPD \) equal (4.), and the side \( DP \) common to both; they are therefore equal, and consequently have \( ED = DF \), and \( EP = PF \), wherefore \( E = FP - Pf = Aa \). 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 \( Ff : fF, or Aa :: FC : CD; \) but \( FC \) is half \( Ff \), therefore \( CD \) is half of \( Aa. \)

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

Prop. XXVI.

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 \( FD, f'd \) perpendiculars from the foci, the rectangle contained by \( FD \) and \( f'd \) 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 Of the Hyperbola is the centre of the hyperbola, and radius CA half the transverse axis; now F'Da 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 f'd, therefore,

\[ DF : df = DF : FH = AF \cdot aF (36. 3. E.) = CB^2 (3.) \]

Cor. If PF, Pf be drawn from the point of contact to the foci, the square of FD is a fourth proportional to fP, FP and CB². For the angles fPd, FPD are equal (4.), and FDP, f'dP are right angles, therefore the triangles FDP, f'dP are similar, and

\[ fP : FP :: f'd : FD :: f'd \cdot FD or BC^2 : FD^2. \]

Prop. XXVII.

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 Db 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, De, which will be perpendicular to the axes.

The triangles DEH, Clm are evidently equiangular, therefore,

\[ DH : DE :: Cl : CM, \] hence \( CL \cdot DH = DE \cdot CM, \) but \( DE \cdot CM \) or \( Ce \cdot CM = BC^2 (10.) \) therefore \( CL \cdot DH = BC^2. \)

In the same way it may be shown, that \( CL \cdot Dh = AC^2. \)

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.

For \( AC^2 : BC^2 :: CL \cdot Dh : CL \cdot DH :: Dh : DH. \)

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

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

\[ GD : DN :: DH : DK; \] and hence \( GD \cdot DK = HD \cdot DN. \)

But \( GD = AC \) (cor. 25.) and \( ND = CL, \) therefore \( AC \cdot DK = HD \cdot CL \) (by the prop.) \( CB^2, \) wherefore \( AC : BC :: BC : DK, \) hence \( DK \) is half the parameter of \( Aa \) (def. 16.)

Definition.

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 the either focus from the centre, and CA the semi-transverse axis, a straight line HG drawn through G, perpendicular to the axis, is called the Directrix of the hyperbola.

Cor. 1. If MFm, 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.

Prop. XXVIII.

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 \( La = Df; \) then CL is evidently half the sum of AL and aL, or of FD and fD, and CE half the sum of FE and fE, and because

\[ Df - DF : Ff :: fE + FE : Df + DF \] (Trig.) by taking the halves of the terms of the proportion,

\[ CA : CF :: CE : CL. \]

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

Cor. 1. If the tangent GMN be drawn through M, the extremity of the ordinate passing through the focus, and 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,

\[ FM : FD :: MO : DK :: GF : GE :: MF : EN, \] therefore \( FD = EN. \)

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 \( ai = aF. \)

Prop. XXIX.

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, PGQ, and the hyperbolic sector PCQ is equal to the hyperbolic trapezium PDEQ.

**Prop. XXX.**

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 K 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,

\[ \frac{CD}{CE} = \frac{CE}{CH}, \]

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 PQG, 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.

**Prop. XXXI. Problem.**

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

**First Method. By a Mechanical Description.**

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

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

If the ruler be made to revolve about the other focus F, while the end of the string is fastened to f, the opposite hyperbola will be described by the moving point D; for in either case Gf—(GD+DF), that is, DF—DF 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. 75. h a k perpendicular to the transverse axis at its extremities, and take AH and AK on each side of the vertex equal to AF, also a h and a k each equal to a F; join H h and K k; take E any point in Aa, and through E draw NE n parallel to HK, meeting H h and K k in N and n. On F as a centre, with a radius equal to EN or En, let a circle be described meeting N n 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. Prop. 28.

**Prop. XXXII. Problem.**

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

Let HA h be the given hyperbola. Draw two parallel straight lines H h, K k 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 D d, and bisect it in E, join CE, meeting the opposite hyperbolas in A and a, then A a will be the transverse axis (5 cor. 8.) for it is perpendicular to D d (3. 3. E.) which is an ordinate to Aa. The other axis will be found by drawing Bb a straight line through the centre perpendicular to Aa, and taking CB so that CB² may be a fourth proportional to the rectangle AE · Ea, and the squares of DE and CA, thus CB is half the conjugate axis (19.). PART IV.

SECT. I. OF THE CONE AND ITS SECTIONS.

Definitions.

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

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

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

Prop. I.

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.

Prop. II.

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 FG, HF be the common sections of the plane EFG, and the triangles VAB, VCD. Because the planes EFG, ADB are parallel, HE, HF will be parallel to CA, CD, and

\[ AC : EH :: (VC : VH ::) CD : HF, \]

but AC = CD, therefore EH = HF. For the same reason GH = HF, therefore EFG is a circle of which H is the centre and EG the diameter.

Prop. III.

If a scalene cone ADBV be cut through the axis by a plane perpendicular to the base, making the triangle VAB, and from any point H, in the straight line AV, a straight line HK be drawn in the plane of the triangle VAB, so that the angle VKH 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 VKH 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 \( HL \cdot LK = EL \cdot LG \) (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.

Prop. IV.

If a cone be cut by a plane which does not pass through the vertex, and which is neither parallel to the base, nor 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 ADVB be any cone, and let ONP be the common section of a plane passing through its vertex and the plane of the base, which will fall without the base, will touch it, or will fall within it.

Let FKM be a section of the cone parallel to VPO; through C the centre of the base draw CN perpendicular to OP, meeting the circumference of the base in A and B; let a plane pass through V, A and B, meeting the plane OVP in the line NV, the surface of the cone in VA, VB, and the plane of the section FKM in LK; then because the planes OVP, MKF are parallel, KL will be parallel to VN, and will meet VB one side of the cone in K; it will meet VA 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 = EL·LG.

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

\[ KL : LG :: KH : HQ, \] and \( HL : LE :: KH : KR; \) therefore \( KL·HL : LG·LE = LF^2 : KH^2 : HQ·KR. \)

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

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 \( DI^2 = AI·IB \),

hence \( DI^2 : FL^2 = AI : IB : EL : LG. \)

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

\[ IB : LG :: KL : KL; \] therefore \( DP : FL^2 :: KL : KL. \)

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

CASE 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

\[ HT : TV :: HL : LE, \] and \( KT : TV :: KL : LG, \) therefore \( HT·KT : TV^2 :: HL·LK : LE·LG or LF^2. \)

Hence it appears, that \( HL·LK \) has to \( LF^2 \) a constant ratio, therefore the section DFKMS is an hyperbola, of which KH is a transverse diameter, and FM an ordinate to that diameter, (2 cor. 19. Part III.).

SCHOLIUM.

From the four preceding propositions it appears, that the only lines which can be formed by the common section of a plane and the surface of a cone, are these five. 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 triangle and circle, may be referred to the hyperbola Of the Curve and the ellipse, for if the axes of an hyperbola be supposed to retain a constant ratio to each other, and, at the same time to diminish continually, till at last the vertices coincide; the opposite hyperbolas will evidently become two straight lines intersecting each other in a point; and a circle may be considered as an ellipse, whose axes are equal, or whose foci coincide with the centre; so that the only three sections which require to be separately considered, are the ellipse, the parabola, and the hyperbola.

SECT. II.

OF THE CURVATURE OF THE CONIC SECTIONS.

DEFINITIONS.

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

LEMMA.

Let PL be any chord in a circle, PX a tangent at fig. so. 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 b 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 Ln.

From G and g draw the straight lines GP, gP, GL, gL, 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

\[ PE : EG :: GN or Eg : NL, \] therefore \( GE^2 = PE · NL. \)

In the same way it may be demonstrated that \( ng = gE, \) and that the triangles PgE, gLn are similar, and therefore that

\[ PE : Eg :: gn or Eg : nL. \]

and hence \( gE^2 = PE · nL. \)

PROP. I.

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.

Let \( Pp \) 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 \( DE \) 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 \( GH, gh, PO \) 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 fig. 82; and in the latter it will fall in \( LP \) produced, as in fig. 82, and fig. 83.

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

Then \( DE^2 = PE \cdot RL \) (Cor. prop. 9 of Part I.)

Now \( GE^2 = PE \cdot LN \)

and \( gE^2 = PE \cdot Ln \) (Lemma).

Now as \( Pp \) 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 \( gh \) 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 \( DE^2 \) 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.

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

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

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

Case I. Let the conic section be a parabola, of fig. 89., of which \( Pp \) is a diameter (fig. 89.) and \( PLK \) the circle of curvature at the vertex, cutting off from the diameter a segment \( PL \) equal to its parameter. Draw \( LK \) a diameter of the circle, and draw \( PO \) perpendicular 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 \( LK \),

\[ IS^2 = PS \cdot PL \] (Lemma);

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

Let \( DE \) an ordinate to the diameter \( Pp \) 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.

\[ DE^2 : GE^2 :: LP : LN :: LO : LH. \]

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

Let the ordinate \( DE \) now meet the arch \( PKI \), anywhere, as at \( g \), draw \( gh \) perpendicular to \( LK \), meeting \( LP \) in \( n \), then it will appear as before that

\[ dE^2 : gE^2 :: LP : Ln :: LO : Lh; \]

but \( LO \) is less than \( Lh \), 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 ellipse or hyperbola (fig. 90.) of which \( Pp \) is a diameter, 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 \( pQ \), a tangent to the conic section in \( p \), 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 \( QP \), \( QL \), meeting \( PL \) in \( S, T \).

Because of the parallels,

\[ PP : PS :: QP : QI :: LP : LT, \]

hence \( PP : LP :: PS : LT :: PS - SP : LT - SP; \)

but \( LT - SP = IS^2 \) (Lemma),

therefore \( PP : LP :: PS - SP : SP; \)

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

Let \( DE \) an ordinate to the diameter \( Pp \) meet the arch \( PLI \) anywhere in \( G \), if the point \( L \) is between... Of the Cur. P and p, or the arch PIL, 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 YV, LQ,

\[ Pp : pE :: (QP : QY ::) LP : LV; \]

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

Prop. I.

\[ DE^2 : GE^2 :: LV : LN :: QY : QZ; \]

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 DE d now meet the other arch PGI anywhere in g; draw gh perpendicular to LK meeting LP in n, and IP in z, then it will in like manner appear that

\[ dE^2 : gE^2 :: LV : Ln :: QY : Qz; \]

and since in this case QY is less than Qz, therefore \( dE^2 \) is less than \( gE^2 \); hence the arch Pgi is wholly without the conic section.

Prop. IV.

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

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

\[ FK : FP :: PH, \text{ or } 4PF : PM, \]

hence \( FK : FP :: FP : \frac{1}{2}PM, \text{ or } \frac{1}{2} \text{ the radius}. \)

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

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

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

\[ \text{for } \frac{2FP^2}{FK} = \frac{2AF \cdot FP^2}{AF \cdot FP \cdot FK} = \frac{L \cdot FP^2}{FK^2}. \]

Prop. V.

The radius of the circle of curvature at the vertex of any diameter of an ellipse, or hyperbola, is a third proportional to the perpendicular drawn from the centre upon the tangent, and half the conjugate diameter; and the chord which is drawn from the point of contact through the focus is a third proportional to the transverse axis, and conjugate diameter.

Let PL be the chord cut off from the diameter, Fig. 91, 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 QC q meeting PH in N and PM in S, then PS is equal to the perpendicular from the centre C upon the tangent. The triangles PSC, PRO are similar, therefore,

\[ PS : PC :: PR : PO, \]

but \( PC : CQ :: CQ : PR \) (Def. of param.)

therefore \( PS : CQ :: CQ : PO. \)

Secondly, the triangles PSN, PHM are similar,

therefore \( PN : PS :: PM : PH; \)

but \( PS : CQ :: (CQ : PO :: ) Qq : PM, \)

therefore \( PN : CQ :: Qq : PH, \)

or, since \( PN = AC \) (Cor. 17. Part II. and Cor. 25. Part III.),

\[ Aa : Qq :: Qq : PH. \]

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

Cor. 2. The radius of curvature is also equal to \( \frac{CQ^2}{AC \cdot BC}, \) for \( PS = \frac{AC \cdot BC}{CQ} \) (15. 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 \( \frac{L \cdot FP^2}{FK^2}. \) For the triangles PFK, NPS are manifestly similar, therefore

\[ FK : FP :: PS : PN, \text{ or } AC :: BC : CQ; \]

hence \( CQ = \frac{FP}{FK} \times BC, \)

and \( \frac{CQ}{AC \cdot BC} = \frac{FP}{FK} = \frac{BC^2}{AC} = \frac{FP^2}{FK^2} \times \frac{1}{2}L. \)

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