4

CONIC SECTIONS.

Atternando, DR x RE: IS x SK:: MR x RN : HS x SG.Therefore, as is obvious from what lias already been shewn,

PR: PS:: DR X RE : IS X SK.

And if S be without the cone, and the line drawn through ittouch the. conic surface instead of cutting it, the reasoning is stillthe same when the square of the tangent is taken in place of therectangle under the segments of the secant.

Prop. VII. Fig. 7, 9.

In the ellipse and hyperbola the rectangle of the abscisses of thetransverse axis, is to the square of their ordinates in a constant ratiowhich is the ratio of the square of the semitransverse axis, to thesquare of the semiconjugate.

Let PQ be the transverse axis’, as in def. 11, and let IG, KO,be ordinates, and LC the senticotijugate; also let the plane VADintersect the vertical plane in the Tine YB, then It). 11. E. PQis narallel to VB, and since IGH is parallel to the base of the cone,vh-: AB x BD :: PG X GQ : IG X GH = IG 3 . Ill the samemanner it mav be shewn that VB-: AB x BD :: POx OQ : KO 2:: PC X CQ'= PC 2 : CL 2 ; therefore PC 2 : Cl. 2 :: PG X GQ :IG 2 : hence the proposition is manifest.

Cor. 1. The square of the transverse axis is to the square of theconjugate as the rectangle of the abscisses to the square of theirordinate.

Cor. 2. Hence at equal distances from the centre, or from thevertices, the ordinates are equal ; for the property is the same onboth sides.

Cor. 3. Hence also the two foci are equally distant from the cen-tre, or from either vertex. v

Cor. 4. Because PG x GQ = PC 2 —GC 2 in the ellipse, andequal GC 2 —PC 2 in the hyperbola, 5, and 6. 2. E. the same pro-perty is thus expressed PC 2 : CL' 2 :: PC 2 v> GC 2 : IG 2 .

Cor. 5. The transverse is to its parameter as the rectangle of theabscisses to the square of their ordinate. For Cor. 1. PQ 2 : 2 CL 22 CL 2

or by division PQ :-:: PG X GQ; or PC 3 t/. GC 2 : IG 2 ;

PQ

2 CL 2

but-is equal to the parameter, def. 17; hence the Cor. is

PQ 2

manifest.

Cor. 6. The distance of each foci of an ellipse from either ex-tremity of the conjugate axis is equal to half the transverse axis ;and the disauce of either of the foci of a hyperbola from the centreis equal to the distance between the extremities of the transverseand conjugate axes.

For let G be the focus, then PC : CL: : CL : IG. def. 17. Cor.therefore PC 2 : C I. 2 :: C L 2 : IG 2

but PC 2 : CL 2 : : PC 2 w GC 2 : IG 2 ,

Cor. 4. hence CL 2 — PC 2 v. GC 2 that is PC 2 — GC 2 in theellipse GC' 2 —PC 2 in the hyperbola; but because GCL is a rightangle, CL 2 = GL 2 — GC 2 ; hence in the ellipse PC 2 =GI. 2 , andPC = GL. And in the hvperbolu GL 2 = 2 GC 2 — PC 2 =GQ 2 +QL 2 + 2 QC x GQ, ( 12 '. 2 . E.) = GC 2 -j-PG x GQ, (6. 2 .E.) = GC 2 -)- 2 QC x GQ + GQ 2 , therefore QL 2 = GC 2 , andQL=GC ; hence the corollary is manifest.

Cor. 7. The semiconjugate is a mean proportional between halfthe parameter and the two segments between the focus and eachvertex of the ellipse, or opposite hyperbolas. For Cor. 6, CL 2

= PC 2 w:GC 2 =PGx GQ.

Prop. VIII. Fig. 8.

In a parabola the abscisses are proportional to the square of theirordinates.

Let the figure be as in the 7th definition, then because EF, AB,are parallel, PG : PQ : : GE : QA, and because GF, QB, areequal, GQ being parallel to FB ; PG : PQ : : GE x GF: QAX QB, but GE x GF = lGx GH = IG 2 , and QA x QB= KQ 2 ; therefore PG : PQ : : IG 2 : QK 2 .

Cor. 1. Hence the third proportion to the absciss, and ordinate,

IG*

or the parameter of the axis is a constant quantity. For-

PG

OK 2

_A_= the parameter of the axis, def. 17.

~ PQ v

Cor. 2. The distance of the focus from the vertex is equal tohalf the ordinate passing tlu ough the focus. For let III be thedouble ordinate passing through the focus, then IG is equal tohalf the parameter, def. 18; then PG : GI : : GI : 2GI, thereforePG = \ GI = J HI = £ of the. parameter.

Cor. 3. The rectangle of the parameter, or four times the dis-tance of the focus from the vertex, and the absciss is equal to thesquare of the ordinate. Thus if G be the focus 4 PG x PQ

= KQ-.

Prop IX. Fig. 17, 19.

Let AB be the transverse, and DE the conjugate axis of anellipse, or hyperbola, or opposite hyperbolas; from any point inthe curve, or opposite curves, as M, let MC be drawn to the cen-tre, and MP perpendicular to the transverse axis, and take CO inthe same axis, such that CO 2 may be equal to MC 2 — CD 2 in theellipse, and to MC* 4- CD 2 in the hyperbola ; then as AC is toCF, so is PC to CO.

For, because AB and DE are conjugate axes, therefore,

AC 2 : CD 2 : : AP x Pi3 : MP 2 (Prop. 7.) therefore, AC 2 : AC 2± CD 2 : : AP x PB : AP x PB + MP 2 . But in the ellipseAC 2 — CD' 1 — CF-; (Prop. 7. Cor. (3.) and AP x PB—MP*= AC 2 - CP 2 - MP* = AC 3 - MC 2 = AC 2 - CD 2 - CO*(hyp.) = CF 2 — CO 2 : and, in the hyperbola, AC 2 -)- CD 2= CF 2 ; (Prop. 7, Cor. 6,)and AP x PB 4-MP 2 = PC 2 - AC*+ MP* = MC 2 - AC 2 = CO 2 - CD* - CA* = CO 1 - CF 2 .Therefore, the last analogy becomes, AC 2 : CF* : : AC 2 i CP*: CF'-ytCO 2 . Consequently, AC 2 : CF 2 :: CP 2 : CO 2 , (19. 5. E.)

And, AC : CF : : CP : CO.

Prop. X. Fig. 17, 19 .

If M be a point in an ellipse or hyperbola, and MF and Nl/bedrawn to the foci; then, in the ellipse, the sum of MF and M/isequal to the transverse axis; and, in the hyperbola, the did'erenceof MF and M/is equal to the transverse axis.

Draw MP perpendicular to the transverse axis, and take CO asin the last prooosition. And, because

‘ AC : CF : : CP : CO, Pr. 9,

Therefore, AC X CO=FC x CP; and 4 AC x CO = 4CFX CP. But, because ABandF/are bisected in C, therefore4 AC x CO = BO 2 - AO 2 , (8. 2. E.) and 4 FC x CP= P/ 2 -PF 2 =f M 2 — MF 2 , 47. 1. E; therefore BO 2 — AO 2 = / M*- MF 2 . J

p*

6 2

Ag in, MF 2 + M./ 2 = /P 2 -f- FP 2 + 2MP 2 = 2FC* +2 C+ 2MP* = 2FC 2 + 2MC* = 2FC 2 ± -CD 2 4- 20= 2AC* + 2CO' 2 = BO 2 4- AO 2 .

And, because BO 2 -(- AO 2 =/M* + MF 2 , and RO 2 — AO*=/ M 2 — MF 2 ; therefore, by adding the equals, 2 RO* = 2/M 2 ;and, by subtracting the equals, 2 M f * = 2 AO 2 . Therefore/M= BO, and FM = AO ; whence the proposition is manifest.

X K.UP, Al.

. i /. lO,

A straight line drawn from any point in a conic section to a focushas to a perpendicular drawn to the corresponding directrix, a ra-tio that is constantly the same wherever the point is assumed inthe curve; and, in the ellipse, the constant ratio is a ratio of mi-nority (or of a less magnitude to a greater) ; in the hyperbola theconstant ratio is a ratio of majority (or of a greater magnitudeto a less) ; and, in the parabola the constant ratio is a ratio ofequality.

Let M, fig. 17 and 19, he a point in an ellipse or hyperbola,and draw MF to a focus, and MK perpendicular to the directrixHG, which corresponds to that focus, draw MP perpendicular tothe transverse axis, and take C O, as in Prop. 9. Then

AC : CF : : CP : CO, Pr. 9.Invertendo, CF : CA : : CO : CP

Therefore, CF : CA: : FO : AP, (19. 5. EA

But, CF : CA : : AF : AG, Def. 19.

Therefore, CF : CA : : AO : GP, (12. 5. E.)

But, as has been shewn in the demonstration of the last nronosi-tion, AO = MF. and GP = MK ; thereforeCF : CA : : MF : MK.

But the ratio of CF to CA is a constant ratio ; and it is a ratio ofminority in the ellipse, and a ratio of majority in the hyperbola.

In the parabola, fig. 18, GA = AF, aud 4 AF x AP = MP 2 ,

Prop.