CONIC SECTIONS. 



the diameter F Q to bisect O S, and, be- 

 cause all the diameters of the curve are 

 parallel, therefore P Q is perpendicular 

 to O S, and an axis of the curve, Def. 17. 

 And because O S can be an ordinate of 

 only one diameter, therefore there is only 

 one* axis. 



Def. 19. Fig. 44, 45, and 46. Let A B 

 Otg.44 and 46.) be the transverse axis, 

 I) E h. conjugate axis, and C the centre 

 of an ellipse, or hyperbola, or opposite 

 hyperbolas : and let C F and C/be taken 

 in the transverse axis, such that C F* and 

 Cf are each equal to C A 2 C D 1 in the 

 ellipse, and to C A.- -f- C D 1 in ihe hyper- 

 bola ; then the two points F and / are 

 called the foci of the ellipse, hyperbola, 

 or opposite hyperbolas. 



But the focus of a parabola (fig. 45.) 

 is a point F in the axis within the curve, 

 and distant from the vertex by a line equal 

 to one-fourth part of the parameter of 

 the axis. 



Cor. The distance of each foci of an 

 ellipse from either extremity of the con- 

 juga e axis is equal to half the transverse 

 axis ; and the distance of either of the 

 foci of a hyperbola from the centre is 

 equal to the distance between the extre- 

 mities of the transverse and conjugated 

 axes. 



Def. 20. If F (fig/ 44 and 46) be a fo- 

 cus of an ellipse, or hyperbola, or oppo- 

 site hyperbolas, and A G be taken in the 

 transverse axis (on the opposite side of 

 the vertex to the focus F,) such, that 

 A F is to A G as C F is to C A ; then a 

 line, as H K, drawn through G perpendi- 

 cular to the transverse axis, is called adi- 

 reetrix of the ellipse, or hyperbola, or 

 opposite hyperbolas. 



Fig-. 45. But the directrix of a parabola 

 is a line, as H K, perpendicular to the 

 axis, drawn through a point G, as far dis- 

 tant from the vertex of the axis on the one 

 side as the focus is on the other side. 



Cor. An ellipse, hyperbola, or opposite 

 hyperbolas, have two directrices ; one 

 corresponding to each focus. For the 

 same construction that is made for one 

 focus may be made for the other focus. 



PBOP. XXIX. 



Fig. 44 and 46. Let A B be the trans- 

 verse, and D E the conjugate axis of an 

 ellipse,or hyperbola, or opposite hyperbo- 

 las; from any point in the curve, or oppo- 

 site curves, as M, let M C be drawn to the 

 centre, and MP perpendicular to the trans- 

 verse axis, and take C O in the same axis, 

 such that C O 1 may be equal to M G 1 



C D> in the ellipse, and to M O+C D in 

 the hyperbola ; then as A C is to C F, se 

 is P C'to C O. 



For, because A B and D E are conju- 

 gate diameters, therefore, 



A C* : C D 1 :: A P X H B : M P 1 , (Cor. 3. 

 Def. 15.) therefore, A C a : A C 2 :p C D 1 :: 

 APxPBiAPxPB^MP^ But in 

 the ellipse A C 1 C D 1 = C F 1 ; and A PX 

 P B M P J A C 1 C P 1 MP a = 

 A C 1 M C' = A C 1 C D 1 C = 

 C F* CO 1 : and, in the hyperbola, A C 1 

 + C D* = C F J ; and A P X P B + M P' 

 = P C 2 A C 1 + M P 1 = M C i 

 A <P = C O 1 C D> C A 1 = C O 1 

 C F-. Therefore the last analogy be- 

 comes, 



A C J : C F 1 :: AC J ^C P 1 : C F^C O 4 

 Consequently, A C 2 : C F 1 :: C P~ : C O* 



19. 5. E. 

 And, AC :CF::C P: C O. 



PROP. XXX. 



Fig-. 44 and 46. If M be a point in an 

 ellipse or hyperbola, and MF and M/be 

 drawn to the foci ; then, in the ellipse, 

 the sum of M F and M /is equal to the 

 transverse axis ; and, in the hyperbola, 

 the difference of M F and M/is equal to 

 the transverse axis. 



Draw M P perpendicular to the trans- 

 , verse axis, and take C O as in the last 

 proposition. And, because 



A C : C F :: C P : C O, Pr 29. 



Therefore, ACxCO = FCxCP; 

 and4ACxCO = 4CFxFO. But 

 because A B and F / are bisected in C, 

 therefore 4ACxCO = BO 1 A O 

 8r2. E. and 4 F C X C P = \ '/> P F i 

 =/M i M FS 47. 1. E ; therefore B O s 

 - A O 1 =/M 2 M F 1 . 



Again, M F 1 + M/ 1 =fP* -\- F P* -f- 

 2 M P 1 =* 2 F C 1 + 2 C P- + 2 M P J = 

 2 F C z + 2 M C 1 = 2 F C 1 2 C D 1 + 

 2CO l = 2AC s -r-2 C O 2 = BO 1 -f- 

 AO 1 . 



Arid, because B O 1 -f A O 1 =/M + 

 M F 1 , and B O 1 A O J =/M a M F 1 ; 

 therefore, by adding the equals, 2 B O 1 = 

 2/M 1 ; and, by subtracting the equals, 

 2 M Y 1 = 2 A O 1 . Therefore /M = B O, 

 and F M = A O ; whence the proposition 

 is manifest. 



PROP. XXXI. 



Fig. 44, 45, and 46. A straight line 

 drawn from any point in a conic section 



