54 MR TALBOT ON CONFOCAL CONIC SECTIONS. 



" If two ellipses and two hyperbolas have all of them the same centre and 

 foci, and therefore intersect in four points, forming a curvilinear quadrilateral, the 

 diagonals of this quadrilateral are equal." 



There will, of course, be a similar quadrilateral on the other side of the axis 

 major. 



In proof of this theorem, it is sufficient to calculate the value of one diagonal, 

 for since that is found to be a symmetrical function of the four greater axes of the 

 given curves, the second diagonal has necessarily the same value. 



This may be shown thus (Vol. XXIII., fig. 15, p. 296). Adopting the former 

 notation, the square of one of the diagonals, or D 2 = (x — xf + (y — y) 2 



where 



x = Aa y = B/> 



x = Aa y = Bb 

 ... D 2 = (a? + y 2 ) + (f + f) - 2- - 2yy 



But xx — kkaa, which is a symmetrical quantity, being the product of the four 



major axes :— and yy = BBbb is a symmetrical also, being the product of the four 



minor axes. 



Therefore it remains to show that (x 2 + y 2 ) + (x 2 + y 2 ) is a symmetrical 



quantity. 



Now 



x 2 + y 2 = A 2 a 2 + B 2 b 2 



but 



B 2 6-' = (1 - A 2 ) (a 2 - 1) = - 1 + (A 2 + a 2 ) - AV 

 .-. x 2 + y 2 = (A 2 + a 2 ) - 1 . 

 And similarly 



x 2 + y 2 = {A 2 + a 2 ) - 1 



.-. (x 2 + y 2 ) +(x 2 + f) = (A 2 + A 2 + a 2 + a 2 ) - 2 



which being a symmetrical quantity, the truth of the theorem in question is 

 demonstrated. 



From this theorem many others may be deduced ; some of which I have given 

 in my first memoir. The following elegant theorem was communicated to me by 

 Charles H. Talbot, Esq. 



" If the direction of one of the diagonals passes through the focus, that of the 

 other diagonal passes through the other focus." 



Demonstration. — First take the general case in which neither diagonal passes 

 through a focus (see fig. 1). Let the diagonals be PP', QQ'; join HP, HP' and 

 SQ, SQ';— then I say that HP'— HP = SQ , -SQ. 



