28 
ON a locus derived from two conics. 
[389 
where 
A 1 2 = (a — by (c — dy, 
B — C =(a+ b)(c + d) — 2 (ab + cd), 
are each of them symmetrical in regard to a, b, and in regard to c, d, respectively. 
Let the equations of the two conics be 
U = (a, b, c, f, g, li\x, y, z) 2 = 0, 
U' = (a', b', c\ /, g', h'Joc, y, z) 2 = 0, 
and let (a, ¡3, y) be the coordinates of the variable point. Putting as usual 
(A, B, C, F, G, H) = (bc—f 2 , ca — g 2 , ab — h 2 , gli — af, hf—bg,fg—ch), 
K = abc — af 2 — bg 2 — ch 2 + 2fgh, 
the equation of the tangents to the first conic is 
(A, B, C, F, G, H%X, Y, Z) 2 = 0, 
where 
X = 7 y — ¡3z, Y = az — yx, Z = fix — ay, 
and therefore 
aX + /3 F+ 7 Z = 0. 
Hence substituting for Z the value — ~(aX + /3Y), we find, for the equation of the 
tangents, an equation of the form aX 2 + 2hXF+bF 2 = 0, which has, in effect, been taken 
to be (X — ciY) (X — bY) = 0; that is, we have 
1 : a + b : ab = a : — 2h : b; 
and, in like manner, if the accented letters refer to the second conic 
1 : c + d : cd = a' : - 2h' : b'. 
Substituting for a, h, b their values, and for a', h', b' the corresponding values, we find 
1 : a + b : ab 
= Ay 2 — 2 Gy a + Ca 2 
: — 2 (By 2 — Fay — G/3y + Ca/3) 
: By 2 — 2F/3y + Cfi 2 . 
1 : c + d : cd 
= A'y 2 -2G'ya + C'a 2 
: — 2 (H'y 2 — Fay — G'fiy + C’a/3) 
: B'y 2 -2F/3y + C'l3 2 . 
We then have 
(a — b) 2 = (a + b) 2 — 4 ab, 
= 4 (Hy 2 — Fay — G(3y + Ca/3) 2 
- 4 (Ay 2 - 2Gya + Ca 2 ) (By 2 - 2F(3y + C/3 2 ), 
= - 47 2 (BC - F 2 , .. .$a, /3, y) 2 , 
= - 47 2 K (a, ... $a, /3, y) 2 ,