20 Mr. A. Cayley on the Porism of the 



we have 



disct.(yt, k, k,a-k, b-k,c-kjx, y, zf= (K, 0, 0', YJJk, If, 



i. e. 



K=-4 



%=±[a + b + c) 



&=-L{a + b-Vcf 



K' = 2«6c, 



and the equation 30^— 4K0'=O becomes 



which is satisfied identically. This is as it should be ; for it is 

 plain that there exists a triangle, viz. the triangle (a;=0, y = 0, 

 r=0), inscribed in the conic U = 0, and circumscribed about the 

 conic V = 0. 



Suppose that the equation of the conic containing the Q is 



9/-4zx=0, 



and the equation of the conic touched by the sides is 



aas'^ + by^ + cz'^ = 0. 



Then the tangential equation of the last-mentioned conic is 



bc^'' + cari^ + ab^=0. 



And if we take for the angles of the triangle a^ :y : z=l ■.2\:X^, 

 or 1 : 2/A : jj,^, or 1 : 2v ; v^, then the equation of the line joining 

 the angles {/u,), (v) is 



2/jbvx — {fi + v)y + z =0, 



which will touch the conic aa;'^ + by'^ + cz'^ = if 



bc4/j,^v^ + ca{/j, + v)'^ + ab .4t = 0. 



And it is required to find under what circumstances the equations 



be . VV + ca(/A + v)2+«^».4 = 



be . 4v2x2 + c«(v + X)2 + «i.4=0 



6c.4\V^ + c«(X + /^)^ + «^.4 = 



become equivalent to two equations only. The condition is of 

 course included in the general formula ; and putting 



disct. {ka, kb + \, kc, 0, -2, QJx, y, zf= (K, 0, 0', IsJJk, If, 



we must have 



302-4K0' = O. 



The discriminant in question is 



k^abc + k^ae - ^ . 4^» - 4 = 0, 



or K = l, 0= ^ae, 0'= — §i, K'= — 4; the required condition 



