296 MR TALBOT ON FAGNANl'S THEOREM. 



Take the equation to the ellipse %+'^=l, and substitute for cc its value Aa, 



And a similar result is obtained from the equation to the hyperbola, viz. — 



A2 B'~ ■ 



.^=1, .-. ^^=a'-l = b' .■.y = Bb. 



For this gives, by putting for a; its value Aa, 



The remainder of the theorem is thus demonstrated, — 

 Since aj = Aa, and j^ = B&, 



= A2a2 + (l-A2) (a2_i) = A2+a2_i=A2 + 62 



= VB2; and therefore CP=VB. 



From heuice we derive the following remarkable property of confocal conies : — 

 If two ellipses and two hyperbolas have all of them the same foci, and inter- 

 sect in four points, k, I, 7n, n, forming a curvilinear 

 quadrilateral, the straight lines /???, kn, which form the 

 diagonals of this quadrilateral, are equal to each other. 

 Demonstration. — Let the semi-axes of the ellipse 

 and hyperbola nearest the centre be called a, h, and A, B ; 

 and let those of the ellipse and hyperbola farthest from 

 the centre be called a', V and A^, B'. 



Let the co-ordinates of the point I be x and y. 



those of m...oc^ ... 2/j 

 those of k ... x^... y.2 

 those of n ••• ^3 ••• 2/3- 

 Then the square of the diagonal Im = {x^ -xf + (y, - yy 

 And of the diagonal ^^ = (^^3 - ^f'-jf + (2/3 - V^f- 



What we have to prove, therefore, is that these two expressions are equal. 



Now, since the point / belongs to the ellipse whose semi-axes are a, b, and 

 also to the hyperbola whose semi-axes are A, B, we have 



x=Aa y —Bb. And for similar reasons, 

 Xj^^A'a' 2/i = B'6' 

 fl72 = Aa' 3/2 = B&' 

 ^3 = A' ay^ = B'b 



Therefore, ^^j = Aa x A' a' 



And. ^2^3 — ^^' ^ -^'^ ' 



And therefore, xxj^=x^x^, each side being the product of the four semi-axes 



major. 



I 



