ON CERTAIN OCTIC SURFACES. 
[654 
88 
© should contain the terms 
2b 2 cf(ch — af) x 6 + 2flx 2 y 2 w 2 , 
viz. in ® the term in x 6 should be — 2b 2 cf(ch — af)af. 
Now writing y = 0, w = 0, we have 
A = cf 2 af, 
[x, y, w) 6 = b 2 f {4 (2 — k) 2 — 2 (6 — 6k + & 3 )} x G , 
= &y(4 — 4& + 2k 2 ) af, 
(x, y, w) 6 = bcf 3 gk 2 x s ; 
and hence the required term of 0 is x 6 multiplied by 
b 2 c 2 fh (4 — 4k + 2k 2 ) 4- 2b 3 cfgk 2 : 
viz. the coefficient is 
= 2b 2 cf[ch (2 — 2k + k 2 ) + bgk 2 ], 
= 2b 2 cf\ch + ch (1 —, k) 2 4- bgk 2 ], 
which in virtue of the relation af+ bgk 2 + ch (1 — k) 2 becomes, as it should do, 
= 2 b 2 cf(ch — af). 
The actual division by (fx 2 + k 2 gy 2 ) 2 would, however, be a very tedious process, and 
it is to be observed, that we only require to know the term 2£lx 2 y 2 w 2 of ©. We may 
therefore adopt a more simple course as follows : the terms of © which contain w 2 
are = (Ax* + 2£lx 2 y 2 + By 4 ) w 2 , hence writing for a moment 
[x, y, w] G = P + Qw 2 , (x, y, w) 6 = R + Sw 2 , 
and observing that we have 
A = c 2 (faf+ gy 2 ) 2 - 2c 2 fg (fx 2 - gy 2 ) w 2 + &c., 
(bx 2 — ay 2 + hw 2 ) 2 = (bx 2 — ay 2 ) 2 + 2h (bx 2 — ay 2 ) w 2 + &c., 
we have 
( fx 2 + k 2 gy 2 ) 2 (Ax 4 + 2ilxHf + By 4 ) = c 2 h (fx 2 + gy 2 ) 2 Q — 2c 2 fgh (fx 2 — gy 2 ) P 
+ (bx 2 — ay 2 ) 2 S + 2h (bx 2 — ay 2 ) R. 
But in this identical equation we may write x 2 = a, y 2 = b, which gives 
(af+ k 2 bg) 2 (Aa 2 + 2£lab 4- Bb 2 ) = c 2 h (af 4- bg) 2 Q — 2 c 2 fgh (af— bg) P ; 
and from the equation 
we have 
P + Qw 2 — (af+ bgk 2 ) 
[x, y, w] G = P + Qw 2 , 
4 {(1 — k) ab + (1 — k) hw 2 } 2 
— 2hw 2 (5 — 6k + k 2 ) ab 
+ 4k 2 (k — 1) bghw 2 (1 — k) ab, 
= — cli (k—1) 2 {4 (k — l) 2 (a 2 b 2 4- 2 w 2 abh) — 2(5 — 6k + k 2 ) hw 2 } 
— 4k 2 (k—1) 2 ab 2 ghw 2 ,