324 
PROFESSOR CAYLEY ON CUBIC SURFACES. 
then the equation in (a, b, c, f, g, h) is that of the circumscribed cone, vertex (a, fi, y, h ) ; 
the order being (as it should be) a'—Q. 
The cuspidal conic is g=0, 4#w-f;s 2 =0, and we at once obtain a 2 —4cg=0 as the 
condition that the line (a, b, c, f, g, h) shall pass through the cuspidal cone. Hence 
attributing to (a, b, c,f, g, h) the foregoing values, we have 
a 2 - 4^=0 
for the eqution of the cone, vertex (a, /3, y, S), which passes through the cuspidal conic ; 
this is of course a quadric cone, d — 2. I proceed to determine the intersections of the 
two cones. 
Kepresenting by 0 = 0 the foregoing equation of the circumscribed cone, and putting 
for shortness 
X=m%f-6h)-2f(2fg+ah), 
I find that we have identically 
0-(/ 2 - bh)X +(g 4 ~ ±ah* - 8 fgh 2 ){a 2 - 4cg) 
— ( 32fg 2 h + 1 d>agh 2 )(af+ bg -+- ch ) = 0 : 
whence in virtue of the relation af-\-bg+ch= 0, we see that the equations 0=0, 
a 2 —4cg=0, are equivalent to 
(f 2 -bh)X.= 0, a 2 -4:cg=0, 
or the twelve lines of intersection break up into the two systems 
f 2 — bh= 0, a 2 — 4cg=0, 
and 
(X = ) 2 7h\f 2 - bh) -2g 2 (2fg+ah)=0, a 2 - 4cg = 0 . 
To determine the lines in question, observe that we have 
( 0) b ~g, a Xa, /3, y, &)=0; 
—h, 0, /, b 
g, -/» 0, c 
— a, —b, —c 
and we can by the first three of these express a. b, c linearly in terms of f, g, h; the 
equations f 2 —bh=0, a 2 — 4cg=0, 27h 2 {f 2 —bh)—2g 2 (2fg-\-ah)=0 become thus homo- 
geneous equations in (f, g, h) ; the equations may in fact be written 
h 2 (a 2 —4cg)= (y 2 + 4a&)$' 2 + fi 2 h 2 —2fiygh — 4@bhf=0, 
l (f-bh) =lf-ah 2 + yhf= 0, 
SX =27 h 2 (ff 2 — odi 2 -f- yhf ) -+- 2g 2 (fih 2 — ygh — 2 Ifg) — 0, 
viz. interpreting (f g, h) 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)3 If- (y 2 -j- 4o% + fiyh} 2 - 4 fi 2 \lf 2 - ah 2 + yhf) 
=(y 2 +4aS){(y 2 4-4a% 2 +/3 2 A 2 — ■ 2fiygh— 4/3 Ihf}; 
