130 
PROFESSOR CAYLEY ON THE CURVES 
tween the parameters (x, y, z, w). Hence, if the conic in addition to the prescribed con- 
ditions passes through two other given points, the point (x, y, z, w ) is given as the 
intersection of a line with the sextic surface; the number of intersections is =6. If 
(x, y, z, w) is situate on the cuspidal curve, then the conic instead of simply touching 
the given conic will have with it a contact of the second order, and if we besides sup- 
pose that the conic passes through a given point, then the point (x, y, z , w) is given as 
the intersection of the cuspidal curve with a plane; the number is =6. Similarly, if 
the conic has two contacts with the'given conic, and besides passes through a given point, 
then the point (x, y, z, w ) is given as the intersection of the nodal curve by a plane ; 
the number is=4. Finally (observing that in the case in question of the contacts of a 
conic with a conic we cannot have three simple contacts, or a simple contact and one of 
the second order), a point of intersection of the nodal and cuspidal curves answers to a 
contact of the third order; and the number is =4. That is, the theory of the sextic 
surface leads to the following values (agreeing with those obtained from the formulae 
by writing therein m=n= 2, «=6), viz. 
(1::) =6, = 2 m+n, 
(1, 1 )=4 , = 2m 2 -\-2mn-\-^n 2 — 2m— \n— fa, 
(2.-.) =6, = a, 
(3:) =4, = — 4m— 3w-f-3a. 
I remark that the section by an arbitrary plane is a sextic curve having 6 cusps and 4 
nodes ; it is therefore a unicursal sextic ; this suggests the theorem that the sextic surface 
is also unicursal, viz. that the coordinates are expressible rationally in terms of two 
parameters ; I have found that this is in fact the case. In doing this there is no loss 
of generality in supposing that a=b=c=l ; and assuming that this is so, and putting 
also — l+h—k, \+h=k l , and therefore 2 h=k-\-Jc^, we have 
A= — 3 kk» 
B = — MjW— (k+k^z, 
C — — x 2 -y 2 —z 2 -\-(k+k 1 )(xy—zw), 
D= 3z(2xy—zw). 
The equation of the sextic surface being, as before, 
A 2 D 2 + 4 AC 3 + 4B 3 D - 3B 2 C 2 - 6ABCD =0, 
I say that tflis equation is satisfied on writing therein 
x +V—\/ sin< P’ 
x —y= \/| (1 +ic a) cos<p, 
2 = 1 , 
w = ^2a — ^ cos 2 <p + ^2a — sin 2 <p, 
