29 
389] ON A LOCUS DERIVED FROM TWO CONICS, 
and similarly 
(c — d) 2 = — 4y-K' (a', ...\a, /3, y) 2 . 
We have, moreover, 
(a + b)(c + d)- 2 (a& + cd) 
= 4 (tfy 2 - Fay - Gßy + Gaß) (Wy- - Fay - G'ßy + C'aß) 
- 2 {Bf - 2Fßy + Gß* ) (A'y* - 2G'ya + G'a 2 ) 
- 2 (B'f - 2F'ßy + G’ß 2 ) (J.y 2 - 2£ya + Ga% 
= - 2f (BC' + jB'O - 2FF, .. ß, y) 2 , 
and substituting the foregoing values, we find 
4(2*+1 fKK'{a, ...$a, ß, yf(a', ...\a, ß, y)* - {(BC + B'C - 2FF, ./3, y) 2 } 2 = 0, 
or putting for shortness 
© = (BC + B'C - 2FF,.,., GW + G'H - AF - JLT,., .£a, /3, y) 2 , 
the equation of the locus is 
4 (2k + 1 )* KK' ,UU'-®* = 0, 
where (a, ß, y) are current coordinates. The locus is thus a quartic curve having 
quadruple contact with each of the conics U = 0, U' — 0; viz. it touches them at their 
points of intersection with the conic © = 0, which is the locus of the point such that 
the four tangents form a harmonic pencil. 
The equation may be written somewhat more elegantly under the form 
4 (2k + l) 2 . KU.K'U' - © 2 = 0 ; 
viz. in this equation we have 
KTJ = (BG £«, /3, y) 2 , 
K'U' = (B'C-Ff ... /3, y) 2 , 
® = (BC' + B'C-2FF, ...$>, /3, y) 2 . 
In the last form the equation is expressed in terms of the coefficients (A, ...), (A', ...) 
of the line equations of the conics, viz. these may be taken to be 
(A, v, ?) s = 0, (A', v, J) ! = 0. 
In particular, if each of the conics break up into a pair of points, viz. (I, m, n) and 
(p, r) for the first conic, (l\ m', n') and (p\ q, r') for the second conic, then the 
line equations are 
2 (Zf + mg + n£) (p£ + qg +r£) = 0 :