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PROFESSOR CAYLEY ON CUBIC SURFACES. 
120. The three transversals are each facultative ; f=b'=3'; t'= 0. 
121. Hessian surface. The equation is 
4«XZ W(3Y +X+ Z + W) -j- Y 2 (X 2 + Z 2 + W 2 — 2XZ — 2XW — 2ZW) = 0. 
The complete intersection with the cubic surface is made up of the lines (Y=0, X=0), 
(Y=0, Z=0), (Y=0, W=0) (the axes) each twice, and of a sextic curve which is the 
spinode curve; <r'=6. 
The spinode curve is a complete intersection 2x3; the equations may in fact be 
written 
Y 3 +Y 2 (X+Z+W)+4aXZW=0, 
3Y 2 +4Y(X+Z+W)+4(XZ-fXW+ZW)=0; 
the nodes D, C, A are nodes (double points) of the curve, the tangents at each node 
being the nodal rays. 
Each of the transversals is a single tangent of the spinode curve ; in fact for the 
transversal Y+Z-|-X=0, W=0, these equations of course satisfy the equation of the 
cubic surface ; and substituting in the equation of the Hessian, we have Y 2 (X— Z) 2 = 0. 
But Y+Z-}-X=0, W=0, Y=Q is a point on the axis W=0, Y=0, not belonging to 
the spinode curve; we have only the point of contact Y+X+Z=0, W=0, X— Z=0. 
Hence /3'=3. 
Reciprocal Surface. 
122. The equation is found by means of the binary cubic, 
aT(T— ?/U) 2 -f (T— ^U)(T— 2lJ)(T— wU), 
viz. writing for shortness 
p=x-\-z-\-w, 
y—xz-{-xw-\-zw , 
o —xzw ; 
this is a binary cubic (#)[T, E T ) 3 , the coefficients whereof are 
3(a+l), —2 ay—p, ay 2 +y, —35, 
and the equation is hence found to be 
+ a 2 {(12y-(3 2 y— (8/3 7 +365)?/ 3 + (30/35+ 8 y % 2 — 36 7 5 j/+275 2 } 
+ 2«{(6y 2 — /3 2 y— 9j35)y 2 +(12/3 2 5— 2j3y 2 — 18y5)y+2y 3 + 275 2 — 9/3y5} 
— (|3y + 1 8/3y5 - 4/3 3 5 - 4y 3 - 2 75 2 ) = 0 ; 
or substituting for j3, y, 5 in the first and last lines their values 
(=x-\-z-\-iv, xz-\-xw-\-zw, xziv), 
