The Conics of Apollonius (1675) - Page 24

and will be equal. And the transverse side (E H) of the figures will be common to both, which is that which is intercepted between the vertices of the sections. Let sections of this kind be called Opposite.

a 4. 6.
b hyp.
c 9. 5.

That each section, D E F and G H K, is a Hyperbola is clear from the 12th of this. Furthermore, with a straight line S A T drawn through A parallel to M N, A S : B Sa

A T : T O, and A S : S Cb

A T : T X. Whence A S sq. : B S * S C (c that is H E * E P)

A T sq. : T O T X (c that is E H H R). Wherefore E P = H R.
### Prop. XV.

Fig. 20.
Fig. 21.

If in an Ellipse, from a point (C) which divides the diameter (A B) into two equal parts, an ordinate line (D C E) is produced on both sides to the section, and it happens that the produced line (D E) is to the diameter (A B) as the diameter (A B) is to another line (D F): the straight line (G H) which is drawn from the section to the produced (D E) equidistant to the diameter (A B), will be able to form a space (D L) adjacent to the third proportional (D F), having as width the line (D H) which is intercepted between it and the section, and being deficient by a figure (M K) similar to that which is contained by the line (D E) to which they are drawn and that (D F) alongside which they are able to be formed. And if (G H) is further produced to the other part of the section (V), it will be bisected by that (D E) to which it had been applied.

a 13. of this.
b 4. 6.
c hyp.
d 1. 6.
e 2. ax. 1.
f 43. 1.
g 5. 2.
h 3. ax. 1.
k 1. 6.
l 4. 6.
m cor. 20. 6.
n hyp.
o 15. 5.
p 23. 6.
q 9. 5.
r 34. 1.
s 14. 6.

Let A N be the line alongside which the applications to A B are able to be formed: and with B N joined, let G X be drawn through G parallel to D E, and through C and X let X O, C P be drawn parallel to A N; and through N, O, P let N R, O S, P T be drawn parallel to A B. It is therefore clear that D C sq.a = rectangle A P. And G X sq.a = rectangle A O. And because A N : C P (A T)b

(A B : C Bc

2 : 1), it will be that T N = A T.d Whence rectangle A P = N P,d and rectangle X Td = Y Te = N S (because T Of = R O). Therefore rectangle A O (G X sq.)g + O Ph = (rectangle N Pd = A Pa = C D sq.)g = H C sq. (G X sq.) + H E H D. And consequently rectangle H E H Dh = rectangle O P (P S S O). Also, H E H D : H L * H Dk

(H E : H Ll

D E : D Fm

D E sq. : A B sq. (n because D E, A B, D F are in continued proportion)o

C D sq.a (P C C A, or P C C B) : C B sq.g

P C : C Bk

P S : S Op :) P S : S O.q (H E H D) : S O sq. Therefore H L H D sq. = (S O sq.)r = H G sq. Which was the first part. Furthermore, with V Q drawn to G X, and Q Z to A N parallel, because A X X Os = (G X sq.r = V Q sq.)r = A Q Q Zs, it will be that A Q : A Xs

(X O : Q Zl

) X B : Q B. Therefore, by dividing, X Q : X A

X Q : Q B.