128 
PROFESSOR CAYLEY ON THE CURVES 
considered as a cubic equation in X, have a pair of equal roots ; or if we write 
A=3c(ab—h?), 
B =(ab— h 2 )w — 2 chz, 
C = —ax‘ l —by‘ 1 —cz 2 -{-2h{xy—zw), 
D=3z(2xy—wz), 
then the required condition is 
A 2 D 2 + 4 AC 3 + 4 B 3 D - 6 ABCD — 3B 2 C 2 = 0. 
Hence the conic 
wX 2 + 2xYZ + 2y ZX + 2zXY = 0 
satisfies the prescribed conditions, if only the parameters (x, y , z, w) satisfy the last- 
mentioned equation, that is, if ( x , y , z, w) are the coordinates of a point on the sextic 
surface represented by this equation. 
The surface has upon it a cuspidal curve the equations whereof are 
A, B, C 
B, C, D 
= 0 ; 
this may be considered as the intersection of the quadric surface AC — B 2 =0 and the 
cubic surface AD — BC=0 ; and the cuspidal curve is consequently a sextic. 
The surface has also a nodal curve made up of two conics ; to prove this I write for 
shortness k=h—\ / ab, k x =h-\-\ / ab ; the values of A, B, C, D then are 
A= — 3 clcJc„ 
B = — kk x w — 2 chz, 
C = 2 — ax 2 — by 2 — cz 1 + 2 li(ccy — zw), 
D=3z(2xy—zw). 
And it is in the first place to be shown that the surface contains the conic 
x:y:z: w=b\/b: b\/ a: 1 : — ^ 2 +| 5 
where 0 is a variable parameter. Substituting these values, we have 
A= — 3 ckJc x , 
B = Jfkft —c(%h-\-*y ab), 
C=2M 1 0 2 -|(3A- X /^), 
D=3(^-|); 
