PROFESSOR CAYLEY ON CUBIC SURFACES. 
325 
and it thus appears that the two conics touch at the points given by the equations 
If 2 — ah 2 + yhf— 0, 
2/3S/-(y 2 +4a%+/3yA=0 : 
we have moreover 
+ (_2S/_ y A)[2 / 35/-(/ + 4a%+|3yA], 
hence at the last-mentioned two points — fih 2 -\-ygh-\-2)fg is =0 ; and the quartic X=0 
thus passes through these two points. 
The conic (a?—cg )-=- 0 and the quartic X=0 intersect besides (as is evident) in the 
point < 7 = 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 
0=-(y 2 -f-4a% 2 +/37i 2 , 
then we have identically 
16/3 % 2 (&/ 2 - ali 2 + yhf) -fl 2 =- [(y 2 + 4 «% 2 + (3 2 h 2 ] + 4/3 2 g 2 (yh + 21 ff 
= — {("/+ 4a% 2 + /3 2 h 2 — 2/3 ygh — 4/3 Ifg } { (y 2 + 4a% 2 + (3 2 h 2 + 2(3ygh + 4/3 Ifg } 
= - l(a 2 —icg) { (y 2 + 4aby/+(3 2 h 2 +2(3ygh + 4/3 Ifg } ; 
and moreover 
2 ( 3((3P-2S/ ? - w A)-Q=(/+W+/BW-2/3 W A-4/3^=S(a«-%). 
Hence when a 2 — -4c^r=0, we have 
r - a h>+ y hf=^ ¥ „ W-W9-vsh =% ; 
and substituting these values in the equation X=0, it becomes 
2W-l§s+2/.|=0, 
viz. multiplying by 16/3 2 S, and omitting the factor O, this is 
27/i 2 n + 16/3fy 4 =0, 
or finally 
16/3fy 4 - 2 7(y 2 + 4 al)g 2 h 2 + 27/3 2 A 4 = 0, 
a pencil of four lines, each passing through the point <7 = 0, h= 0, and therefore inter- 
secting the conic 
(y 2 -{- 4 ah)g 2 -\- (3 2 h 2 —2/3 ygh— 4/3 &/i/= 0 
at that point and at one other point ; and we have thus four points of intersection, which 
are the required four points. 
Recapitulating, the conic « 2 — -4cy=0 meets the sextic (f 2 —hhfL — 0 in the two points 
fJ/’-rf+y7/=0, 
l2/35/-( y ’+4a%+/3//i=0 
