CONIC SECTIONS. 



in all positions of the lines P Q, and M N, 

 whether they cut, or touch, the same or 

 opposite surfaces. 



PROP. VIII. 



Fig. 10. If a right line, as P T, drawn 

 through a point P in the surface of a 

 cone, so as to be parallel to a right line 

 V B contained in the conic surface meet 

 two parallel lines (in the points R and S) 

 that cut or touch the conic surface or 

 opposite surfaces : then P R is to P S as 

 the rectangle under the segments of the 

 secant, or the square of the tangent, 

 drawn through the point R, is to the rec- 

 tangle under the segments of the secant, 

 or to the square of the tangent drawn 

 through the point S. 



Through the two parallels P T and V B 

 (fig. 10.) draw a plane cutting the conic 

 surface again intheline V A, and the plane 

 of the base in the line B A ; ,and, through 

 R and S, draw M N and H G parallel to 

 A B. Because P T is parallel to V B, 

 and R N to S G, therefore R N G S is a 

 parallelogram ; and R N is = G S. It is 

 obvious that the triangles P M R and 

 P H S are equiangular : therefore P R is 

 to P S as M R is to H S, 4. 6. E, or as 

 M R x R N is to H S X S G, 1. 6. E. But 

 M R X R N and H S X SG are respect- 

 ively equal to the rectangles contained 

 by the segments of any two lines, parallel 

 to the base of the cone, drawn through 

 R and S to cut the conic surface, Cor. 

 Pr ; 5, and hence the proposition is mani- 

 fest, when P T meets two lines parallel 

 to the plane of the base. 



And if P T meet two parallel lines D E 

 and I K, not parallel to the plane of the 

 base ; then let the same construction be 

 made as before : and because D E is pa- 

 rallel to I K, and M N to G H ; therefore, 



DRxRE:MRXRN::ISxSK: 

 HS X S G; 



Alternando, D RxRErlSxSK:: 

 M R x R N : H S x S G. Therefore, as 

 is obvious from what has already been 

 shewn, 



PR:PS::DRXRE:ISXSK. 



And if S be without the cone, and the 

 line drawn through it touch the conic 

 surface instead of cutting it, the reason- 

 ing is still the same, when the square 

 of the tangent is taken in place of the 

 rectangle under the segments of the se- 

 cant. 



PROP. ix. 



Fig. 11. Let a scalene cone be cut by 

 a plane drawn through the axisperpendi- 

 culiar to the plane of the base, making the 

 triangular section V A B ; and let V D, 

 cutting A B produced in D, be drawn so 

 as to make the angle B V D equal to the 

 angle V A B, and draw M N in the plane 

 of the base, perpendicular to A D ; then 

 every section of the cone, as P S Q, made 

 by a plane parallel to the plane V M N 

 (called a subcontrary section) is a circle; 

 and every circular section of the cone, 

 which is not parallel to the base, is a sub- 

 contrary section. 



Draw T S in the plane of the section 

 parallel to M N, which is plainly possible, 

 because the two planes P Q and V M N 

 are parallel : because T S is parallel to 

 M N, a line in the plane of the base, there- 

 fore every plane drawn through S T will 

 cut the base in a line parallel to S T (16. 

 11. E.:) therefore L O K, the common sec- 

 tion of the base, and a plane drawn 

 through V and S T is parallel to S T and 

 M N 1 9. 11 E.) : therefore K O L'is per- 

 pendicular to A B, and it is bisected in 

 O ; therefore S T is bisected in R. Again, 

 the line P Q is parallel to V D, therefore 

 VD Z :PRXRQ::ADXDB:TRX 

 R S (6.) But if a circle be described 

 about the triangle A V B, D V will be a 

 tangent of that circle (32. 3. E) : there- 

 fore V D 2 = A D X D B, and consequent- 

 ly PRxRQ =T R x R S, or R T 1 (36. 

 3. E.) Because the plane A V D is per- 

 pendicular to the base (h y />) and M N is 

 perpendicular to A D : therefore M N is 

 perpendicular to the plane A V D : there- 

 fore, T R, parallel toM N, is perpendicu- 

 lar to the same plane, and to P Q. And 

 hence,from what has already been shewn, 

 the section P Q is a circle. 



Next, let P Q be a circular section, not 

 parallel to the base of the cone: draw a 

 plane through the vertex, parallel to the 

 plane P Q, and let it cut the base in the 

 line M N : draw A D through the centre 

 of the base perpendicular to M N, and let 

 a plane drawn through V and A D cut the 

 parallel planes in the lines P Q and V D, 

 and the conic surface in the lines A V and 

 V B : draw the plane V T L K S through 

 S T parallel to M N, as before. It is shewn, 

 as above, that T S is bisected in R : and, in 

 like manner, it may be proved that any 

 other line, as G H, parallel to M N, is bi- 

 sected. Because P Q, a line in a circle, 

 bisects two or more parallels, it is a di- 

 ameter of the circle, and it cuts all the 



