654] ON CERTAIN OCTIC SURFACES. 89 
that is, 
P = — 4ch(Jc—iy a 2 b 2 , 
Q = (Jc — l) 2 obli [ch (— 6k 2 + 4>k + 2) — 4k 2 b\, 
whence 
(k — l) 2 (Aa? + 2D.ab + Bb 2 ) = ab (af + bg) 2 [ch (— 6k 2 + 4& + 2) — 4A; 2 ò^] 
+ 8 fg («/- %) (& -1) 2 (^)' 2 - 
But we have 
J.a 2 -f Bb 2 = — 2ab (af — bg) (— q/% + c 2 h 2 + 2chaf+ 2chbg), 
and thence 
2 (& — l) 2 il = (0/+ %) 2 [ch (— 6& 2 + 4>k + 2) — 4>k 2 bg] 
+ (k — l) 2 (af — bg) (— 2afbg + 2c l h? + 4<chaf-\- 4chbg\ 
\+ Safbg ) ’ 
or 
(k — l) 2 il = (af + bg) 2 [ch (— 3k 2 + 2k + 1) — 2k 2 bg] 
+ (k — l) 2 (af— bg) [3 afbg + c 2 h 2 + 2chaf-\- 2 chbg]. 
Writing — Sk 2 + 2k + 1 = — 3 (k — l) 2 — 4 (k — 1), this is 
4- (cf— bg) [Safbg + c 2 h 2 + 2chaf+ 2chbg], 
or since 1 : —k : k—l=\ : fi : v ; and writing for shortness (af bg, ch) = (a, /3, 7), 
this is 
ft = (tt 4- BY jy ^ "t ( a — /5) + 7 2 27a + 27/3}, 
which is the value of ft: viz. the conclusion arrived at is that, eliminating 6 from 
the equations 
( f6 2 + g) (kdy — x) 2 + h(k — l) 2 6 2 z 2 = 0, 
(bk 2 6 2 — a)( 6y — x) 2 + h (k — l) 2 6 2 w 2 = 0 : 
where k denotes a determinate function of af bg, ch, viz. writing af bg, ch = a, /3, 7 
and 1 : — k : k — 1 = X : g : v, we have 
A + g + v — 0, 
aA 2 + fig 2 + yv 2 = 0, 
equations which serve to determine k : the result of the elimination is the octic 
equation 
b 2 c 2 f 2 x? + ... + 2 Qx 2 y 2 z 2 w 2 = 0, 
XI = (af + bg) 2 
ch 
(-3- 
4 
k-1 
2k 2 
(k-iy 
bg 
where ft has the last-mentioned value. 
C. X. 
12