454 
A MEMOIR ON CUBIC SURFACES. 
[412 
and we can by the first three of these express a, b, c linearly in terms of / g, h; 
the equations f 2 — bit = 0, a 2 — 4eg = 0, 27/i 2 (/ 2 — bh) — 2g 2 (2fg 4- ah) = 0 become thus 
homogeneous equations in (/ g, h); the equations may in fact be written 
8 2 (a 2 — 4c#) = (7 2 + 4a8) g 2 + /3 2 h 2 — 2/3ygh — 4 /38// = 0, 
8 (/ 2 — bh) — 8f 2 — ah 2 4- 7/«/= 0, 
8X = 27A 2 (8f 2 — ah 2 4- 7/«/) + 2# 2 (/3/f- — >ygh — 28fg) = 0, 
viz. interpreting (/ <7, /¿) as coordinates in piano, the first equation represents a conic, 
the second a pair of lines, and the third a quartic. 
We have identically 
{2/38/— (7 2 4- 4a8) g 4- ¡3yh} 2 — 4/3 2 8 (8f' 2 — ah 2 4- 7hf) 
= (7 2 + 4aS) K7 2 + 4a8) g 2 + /3 2 h 2 — 2(3ygh — 4/38//) ; 
and it thus appears that the two conics touch at the points given by the equations 
8/ 2 — ah 2 4- 7//= 0, 
2/38f— (ry 2 4- 4<a8)g + ¡3<yh = 0 : 
we have moreover 
— (7 s 4- 4a8) (/3h 2 — 7gh — 2 8fg) = 4/38 (8/ 2 — aA 2 4- 7//) 
+ (- 28/- yh) [2/38/- ( 7 2 4- 4ta8)g + /3yh\, 
hence at the last-mentioned two points — (3h 2 4- ygh + 28fg is = 0 ; and the quartic X = 0 
thus passes through these two points. 
The conic (a 2 — cg) = 0 and the quartic X = 0 intersect besides (as is evident) in 
the point g = 0, h= 0 reckoning as two points, since it is a node of the quartic; and 
they must consequently intersect in four more points: to obtain these, in the most 
simple manner, write for a moment 
H = — (7 2 4- 4a8) g 2 4- /3 2 h 2 , 
then we have identically 
16/3 2 8g 2 (8f 2 — ah 2 4- yhf) — H 2 = — [(7 2 4- 4a8) g 2 4- /3 2 h 2 ] 4- 4/3 2 / (yh 4- 28/) 2 , 
= — {(7 2 4- 4a8) g 2 + ¡3rh 2 — 2(3ygh — 2(38fg) {(y 2 4- 4oc8) g 2 4- /3 2 h 2 4- 2(3ygh 4- 4/38//, 
= — 8 (a 2 — 4c^r) K7 2 4- 4aS) g 2 4- (3 2 h 2 4- 2f3ygli 4- 4f38fg]; 
and moreover 
2/3 (/3h 2 — 28fg — ygh) — il — (7 2 4- 4a8) g 2 4- /3 2 li 2 — 2/3ygh — 4(38fg = 8 (a 2 — 4eg). 
Hence when a 2 — 4c# = 0, we have 
il 2 n 
8/ 2 - ah 2 4- 7hf= iopfip > № ~ W9 ~ 79 h = Jp 5