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 less than the greater 

 semidiameter 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, and it will 

 be without the 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 

 D 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 GH, 

 drawn to bisect the angles comprehended 

 by the equal diameters D I and E F, will 

 be conjugate diameters and axes of the 

 hyperbolas. The demonstration is the 

 same as for the ellipse. 



PROP. xxvi. 



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

 lipse are always unequal ; and the greater 

 axis 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. 



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

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

 any semidiameter; draw HP perpendicu- 

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

 Because A B and D E are conjugate dia- 

 meters ; and H P an ordinate to A B, and 

 \l Q. an ordinate to D E : therefore, 



A B 1 : D E 1 :: A P X P B :: H P J , 

 Cor. 3, Def. 15. 



Now, if A B be supposed to be equal to 

 D E, it will follow that A P X P B = HP J ; 

 therefore APxPB-f-CP 1 = HP i -f 

 C P J , or A C*=C H 1 . Therefore, A C = 

 C H: and the ellipse will be a circle, which 

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

 ^nd D E are unequal : let A B be suppos- 

 ed to be greater than D E. 



Because A B s is greater than D E 1 , 

 therefore A P X P B is greater than H P J ; 

 andAPxPB-fCP V>r A C-,\s greater 

 than H P- +PC J ,or C H J . Therefore the 



semi-axis A C is greater than any other 

 semi-diameter H C. 



In like manner. 

 D M* : A B' :: D Q X Q E : H Q>. 



Therefore D Q x QEis less than HQ 1 ; 

 and DQ x Q E 4- C Q*, or C D 1 , is less 

 than H Q* + C Q 1 , or C H 1 . Therefore 

 the semi-axis D C is less than any other 

 semidiameter C H. 



Fig. 42. In the hyperbola, a tangent of 

 the curve drawnfrom the extremity of the 

 axis C A, at A T, falls between the centre 

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

 mi-axis, is less than any other line drawn 

 from C to A T, much more is it less than 

 a semidiameter 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 less, 

 the conjugate axis ; and, in the hyperbo- 

 la, that one is the transverse axis which is 

 a transverse diameter, and the other is 

 the conjugate axis. 



PROP. XXVII. 



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 ; join HK, and draw 

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

 C L to bisect H 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 M C is greater than the angle G M C, 

 (25. I.E.) that is, the angle K L C is 

 greater than the angle H L C. And be- 

 cause H 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, therefoie the angle A M C, or K L C, 

 is less than the angle G M C, or H L C. 

 Therefore C K is less than C H. 



PROP. XX 1 III. 



Fig. 43. A parabola has only one axis. 

 Let O S, terminated by the curve, be 

 perpendicular to any diameter, anddraxv 



