CONIC SECTIONS. 



or the rectangle under the segments of 

 the secant, is greater than the square of 

 the semi-diameter parallel to the tangent 

 or secant. 



First, let the point P be without the 

 hyperbola, and within the asymptotes, 

 and let P H, (fig. 35.) parallel to the 

 semi-diameter C D, touch the hyper- 

 bola; because P is a point within the 

 asymptotes, the line drawn from it 

 through the centre will be a transverse 

 diameter : thus, 



CE* : CD 1 :: EP X PK : PH 1 (19 

 and Cor. 1. Def. 15.) 



But C E 1 is greater than E P X P K ; 

 therefore C D 1 is greater than P H 1 . And 

 in like manner may the proposition be 

 demonstrated, when the line drawn from 

 P does not touch, but cuts the hyperbola, 

 or opposite hyperbolas. 



Next, let P be without the asymptotes : 

 draw R S (fig. 36), terminated by one of 

 the hyperbolas parallel to C P, the line 

 drawn from P to the centre : draw the 

 diameter C E to bisect R 8 and M N 

 through P parallel to C E. Because the 

 diameter C E bisects R S parallel to C P, 

 therefore M N, parallel to C E, is ordi- 

 nately applied to the second diameter C P 

 (18.). Let C Q. be the magnitude of this 

 semi-diameter, then 



C Q 1 : C E :: C Q 1 -f C P* : P N (20) 

 And C E^ : C I) 1 :: M P x P N or P N 1 

 : PH* 



Ex aequo, C Q 1 : C D* :: C Q J -f-C P : P H 1 . 

 But C Q 1 -f C P 2 is greater than C Q 1 , 

 therefore P H 1 is greater than C D 1 . And 

 in like manner may the proposition be 

 proved, when the line drawn through P 

 does not touch, but cuts a hyperbola, or 

 opposite hyperbolas. 



PHOP. XXIY. 



Fig. 37, 38. If from a point (P or Q) 

 in an asymptote of a right line be drawn 

 to touch or cut the hyperbola, or oppo- 

 site hyperbolas (P H or Q R S) : the 

 square of the tangent, or the rectangle 

 under the segments of the secant (P H 2 

 or Q R X Q S) is equal to the square of 

 the semi -diameter (C D 1 ) parallel to the 

 tangent or secant. 



For if not, make H O andRO' X O' S 

 equal to C D 1 : then O and CM are with- 

 out the hyperbola, and they must be 

 either within the asymptotes or without 

 them. In the former case HO 1 and ROi 



X O S would be less than C D 1 (23.) ; 

 and in the latter case H O z and R O X 

 O 1 S would be greater than C D 1 (23.) 

 which are equally absurd. Therefore P H 1 

 and Q R x Q S are equal to C D\ 



Cor. 1. If a tangent of a hyperbola meet 

 both asymptotes, as P M, the segments 

 P H and H M between the asymptotes and 

 the point of contact are equal. And if a 

 right ime cut a hyperbola, or opposite 

 hyperbolas, and botli the asymptotes, as 

 Q T, the segments between the curve or 

 curves and the asymptotes are equal to 

 one another : that is Q R = T S, and Q S 

 = T R. 



For P H 1 , H MS Q R X Q S, and T R X 



T S, are all equal to C D*. 



Cor. 2. On the contrary, if P M, in- 

 tercepted between the asymptotes, meet 

 the hyperbola in H, and is bisected 

 there; then P M is a tangent of the hy- 

 perbola. 



Cor. 3. If any number of lines all paral- 

 lel to one another, as Q, T and M N, cut 

 a hyperbola and the asymptotes, the rec- 

 tangles Q R x R T, M L X L N, under 

 the segments between the curve and the 

 asymptotes, are all equal. 



For it is plain (Cor. 1.) that the rectan- 

 gles are all equal to C D 2 . 



Def. 17. A diameter of a conic section 

 that cuts its ordmates at right angles is 

 called an axis. 



Cor. Because two conjugate diameters 

 of an ellipse, and opposite hyperbolas, cut 

 their ordinates in the same angles, Pr. 

 18; therefore, if there be one axis of 

 these curves, there will necessarily be 

 two, and these will be conjugate diame- 

 ters, and they will cut one another at 

 right angles. 



PROP. xxv. 



Fig. 39,$0. An ellipse, and opposite 

 hyperbolas, have two axes. 



Find the centre of the ellipse C(fig. 39.) 

 and draw two diameters. 



Then if the two diameters be equal to 

 one another, as E F and D I, two other 

 diameters, A B and G H, drawn to bisect 

 the angles contained by D I and E F, will 

 be axes of the ellipse. Join E D and D F: 

 then the lines A B and G II, which bisect 

 the vertical angles of the isosceles trian- 

 gles F C D and E C D, will bisect the ba- 

 ses D E and D F, and likewise cut these 

 lines at right angles. Hence it is plain 

 that A B and G H are conjugate diajne- 

 ters and axes of the ellipse. 



