CON T !C SECTIONS. 



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

 cause aH 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. 



J)ef. 19. Fig. 44, 45, and 46. Let A B 

 (fig 1 . 44 and 46. ) be the transverse axis, 

 D E the conjugate axis, and C the centre 

 of an ellipse, or hyperbola, or opposite hy- 

 perbolas : and letC F andC/be taken in 

 the transverse axis, such that C F 1 and 

 C /> are each equal to C A 1 C D 1 in the 

 ellipse, and to C \- -\- C D 1 in the hyper- 

 bola ; then the two points F and/are call- 

 ed 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 aline 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- 

 jugate 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 a di- 

 rectrix 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 cor- 

 responding 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 cur' es, asM, letMC be drawn to the 

 centre-, andMP perpendicular to the trans- 

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

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



C D 1 in the ellipse, and to M C'-fCD 1 in 

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

 is P C to C O. 



For, because A B and D E are conju- 

 gate diameters, therefore, 



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

 Def. 15.) therefore, A O : A C- q: C D 1 :; 

 A PX PB: AP X PB^MP 1 . But in 

 the ellipse A C J C D S =C F-; and A P X 

 P B - MP Z = AC 1 CP* MP 1 =3 

 A C 1 M C 1 = A C C D> C O 1 =r 

 C F 1 C O : and, in the hyperbola, A C 1 

 + C D 1 = CF l ; andAP x PB-f MP 

 = P C 1 A C* + M P 1 == M C 1 

 A C 1 = C O C D 2 C A> = C O* 

 C F 1 . Therefore the last analogy be. 

 comes, 



AC 1 : CF'-AC'rpCP'iCF'qFCO 1 



Consequently, A C' : C F 1 :: C P 1 .- C O 1 



19. 5. E. 

 And, A C : C F :: C P : C O. 



PHOP. xxx. 



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

 ellipse or hyperbola, and M F and M/be 

 drawn to the foci ; then, in the ellispe, 

 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 pro- 

 position. And, because 



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



Therefore, ACxCO = FCxCP; 

 and 4 A C x C O = 4 C F X F O. But 

 because A B and F / are bisected in C, 

 therefore 4AC x C O = B O A O, 

 8. 2. E. and 4 F C x C P = P/ 1 P F 2 

 =/ M 2 M F 2 , 47. 1 E ; therefore B Q- 

 AO' = /M 2 MF 2 . 



Again, M F J + M/ 2 =/P J -f F P l + 

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

 2FC i -f-2MC J =2FC 2 2CD J + 

 2CO'==2AC i 4-2C O 1 = B O> + 

 AO'. 



And, because B O 1 + A O 2 = /M 2 + 

 M F 1 , and B O 1 A O 1 =/M 2 M F ; 

 therefore, by adding the equals, 2 B 0*= 

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

 2 M F 2 = 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 



