CONIC SECTIONS. 



But if the two diameters be not equal, 

 as M N and P Q, describe a circle from the 

 centre C with a radius le than the greater 

 semidiumetcr C M, but greater than the 

 less semidiameter C P : then the circle- 

 will cut the diameter M N on both sides 

 of the centre within the ellipse, am) it will 

 be without tho ellipse towards the point 

 P ; therefore the circle will cut the pe- 

 riphery of the ellipse both between P 

 and M, and between P and N : let E and 

 1) be the points of section ; then two di- 

 ameters drawn through them will be 

 equal, and the axes of the ellipse will be 

 found as above. 



In the case of opposite hyperbolas, (fig, 

 39.) find the centre C, and from C as a cen- 

 tre describe a circle through a point with- 

 in one of the hyperbolas : then that circle 

 will cut the hyperbola in two points D 

 and E, and two transverse diameters 

 drawn through these will be equal to one 

 another ; and two diameters A B and G H, 

 drawn to bisect the angles comprehended 

 by the equal diameters D I and EF, will be 

 conjugate diameters and axes of the hy- 

 perbolas. The demonstration is the same 



for the ellipse. 



-Fig-. 41 and 42. The two axes of an el- 

 iipse are always unequal ; and the greater 

 -ixis is the greatest diameter, and the less 

 axis the least diameter, of the curve. And 

 that axis of a hyperbola, which is a trans- 

 verse diameter, is the least of all the trans- 

 verse diameters. 



\ B and D E (Fig. 41.) be the two 

 s of an ellipse, C the centre, and C H 

 any Bemidiuneter ; draw H P perpendicu- 

 lar to A H, and II Q perpendicular to D E. 

 Because A I? and 1) E are conjugate dia- 

 meters ; and II P an onlinate to A B, and 

 H Q an ordinate to D E ; therefore, 



AB : I) 1. A P x HB .: IIP', 

 Cor. 3, Def. 15. 



Now, if A B be svipposed to be equal to 

 T) E, it will follow that A P X P B= 1 1 I' ; 

 therefore, APxPB-fCP l =HP--f 

 < \>2, r A C2= C 112. Therefore, A C = 

 <' H : and the ellipse will bo a circle, which 

 is not the case, Cor. 9. Therefore, A B 

 Mid 1) K are unequal ; let A B hi- supposed 

 to be greater than I) K. 



lit cans- A R2 is greater than D E*, 

 therefore A Px P Btt greater than II !'. , 

 .'iid A P x P B -4- C P2, or A t*2, is greater 

 thahHPS-f PrV>rr H' Th-j- fore the 



semi-axis A C is greater than any othe; 

 semidiameter H C. 



In like manner. 



I) M2 : AB2 :: D Q X Q E : H Q 2 . 



Therefore D Q X Q E is less than M (I- . 

 and D Q X Q E -H C Q2, or C L>2, is k-ss 

 than II Q2 + C Q2, or C H^. Therefore the 

 semi-axis 1) C is less than any other semi- 

 diameter C H. 



Fig. 42. In the hyperbola, a tangent of 

 the curve drawn from the extremity of the 

 nxis C A, as A T, falls between the centre 

 and the curve ; and because C A, the semi- 

 axis, is less than any other line drawn from 

 C to A T, much more is it less than a semi- 

 diameter C H drawn from C to the curve 

 on the other side of A T. 



Cor. Hence it is plain, that an ellipse, 

 or opposite hyperbolas, have only two 

 axes. 



Def. 17. The greater axis of an ellipse 

 is called the transverse axis ; and the lss, 

 the conjugate axis ; and, in the hyperbola, 

 that one is the transverse axis which is 

 a transverse diameter, and the other is 

 the conjugate axis. 



PROP. TXTII. 



Fig. 41 and 42. A diameter of an el- 

 lipse nearer the transverse axis is greater 

 than one more remote ; and a transverse 

 diameter of the hyperbola nearer the 

 transverse axis is less than one more re- 

 mote. 



Let C K and C H (fig. 41.) be two semi- 

 diameters of an ellipse j join H K, and draw 

 A G parallel to II K, join C G and draw 

 C L to bisect II K. Because C L bisects 

 H K, it will likewise bisect A G. Cor. 14. 

 And because A M = M G, and A C is 

 greater than C G, therefore the. angle 

 A \I < is greater than the angle G M < . 

 (25. 1 E." that is, the angle K L C is 

 greater than the angle H L C. And be- 

 cause II L = L K, therefore K C, nearer 

 to C A, is greater than H C, more remote 

 from C A, 24. 1. E. 



In the hyperbola, the same construction 

 being made, because A C is less than C G, 

 therefore the angle A MC,or KLC.isless 

 than the angle G M C, or H L C. There- 

 for- <; K is less than C H. 



PHOP. XXVIII. 



Fig. 43. A parabola has only one a\i- 

 Let O S, terminated by the curve, be 

 perpendicular to any diameter, and draw 

 R 





