CONIC SECTIONS. 



of the cone, drawn through the point P, 

 to cut the conic surface. 



Fig. 6. Let the point B be without the 

 base of the cone, and let QR, drawn 

 through P, without or within the conic 

 surface, be parallel to V B, and let it cut 

 the conic surface in Q and R : through P 

 and the line V B draw a plane cutting 

 the conic surface in the lines V G and 

 V H, and the plane of the base in the 

 line B G H ; and through P draw L K 

 parallel to G H. Because V B and PRO, 

 are parallel, therefore the line P R Q, is 

 contained in the plane B V P, 7. 11. E ; 

 and the points Q and R are in the lines 

 V H and V G, the common sections of 

 the plane and the conic surface. Be- 

 cause Q P is parallel to V B, and L K to 

 GH, therefore the triangle QPL is equi- 

 angular to the triangle VBH, and the 

 triangle P K R to the triangle V G B : 

 therefore 4. 6. E. 



VB:PR::BG:PK 

 VB:PQ::BH:PL 



Consequently, V B 1 : PR X P Q. " B G 

 X B H : P K X P L, 23. 6. E. But the 

 rectangle B G X B H is equal to the 

 rectangle under the segments of any 

 other line drawn from B to cut the base of 

 the cone, 35, and 36. 3. E ; and the rectan- 

 gle PKxKL is equal to the rectangle 

 under the segments of any other line, par- 

 allel to the plane of the base, drawn from 

 P to cut the conic surface, Cor. Pr. 5 ; 

 and hence the proportion is manifest in 

 this case. 



Fig. 7. And if the point B be within 

 the base of the cone, and a straight line 

 as (PQR), parallel to the line V B that 

 joins the point B and the vertex of the 

 cone, be drawn to cut the opposite sur- 

 faces through a point P, situated without 

 or within the cone : the proposition may 

 be demonstrated, in this case, in the very 

 same words as in the former case. 



And if the point P (fig. 8.) be without 

 the cone as well as the line V B, and 

 PS, parallel to V B, be drawn to touch 

 the conic surface, instead of cutting it ; 

 then the plane PVB will meet the conic 

 surface in a line V S M ; and B M will 

 touch the base of the cone, and P N, 

 parallel to B M, will touch the conic 

 surface. And because the two triangles 

 S P N and V B M are equiangular, 

 therefore, 



But BM 1 ia equal to the rectangle un- 

 der the segments of any line drawn from 

 B to cut the base of the cone ; and P N* 

 is equal to the rectangle under the seg- 

 ments of any line, parallel to the base of 

 the cone, drawn from P to cut the conic 

 surface ; and hence the proposition is 

 manifest in this case also. 



Fig. 9. If a point be assumed without 

 or within a cone, and two lines be drawn 

 through it to meet a conic surface, or op- 

 posite surfaces, and so as to be parallel to 

 two straight lines given by position ; then 

 the rectangle under the segments of the 

 secant, or the square of the tangent, pa- 

 rallel to one of the lines given by position, 

 has to the rectangle under the segments 

 of the secant, or to the square of the 

 tangent, parallel to the other line given 

 by position, a ratio that is constantly the 

 same, wherever the point (from which 

 the lines are drawn) is assumed, without 

 or within the cone. 



Let V B and V C be two straight lines, 

 (fig. 9.) drawn from the vertex of a cone 

 to the plane of the base, and given by po- 

 sition (.or parallel to lines given by posi- 

 tion :) and let P Q and M N be two 

 straight lines drawn through any assumed 

 point, as R, to cut the conic surface, and 

 so as to be respectively parallel to C V 

 and VB : and as CV 1 is to the rectangle 

 CK X CL (contained by the segments 

 of any line drawn from C to cut the base 

 of the cone,) so let D, any assumed line, 

 or magnitude be to E ; and as V B 1 is to 

 B G X B H (the rectangle contained by 

 the segments of any line drawn from B 

 to cut the base of the cone,) so let F be 

 to E ; and draw S T parallel to the base 

 of the cone through the point R ; then, 

 Pr. 6. 



(C V J : C K X C L, or) (D : E :: P R X 

 RQ : S R X RT, and B V 1 : B G 1 X BH, 

 or) F : E :: M R X R N : S R X R T. 



Therefore, invertendo and ex aequo. 



D : F :: PR X 



= MR X RN. 



VB : PS :: BM :IPN 



And 



And, as the same reasoning applies where- 

 ever the point R is assumed, therefore the 

 ratio of the rectangles P R X R Q. and 

 MR X RN, is the same with, or equal to, 

 the constant ratio of D to F, wherever 

 the point R is assumed. 



And in like manner may the proposi- 

 tion be demonstrated in all other cases, or- 



