402 
ON A SURFACE OF THE EIGHTH ORDER. 
[526 
and if these planes have a common line, the point (x, y, z, w) is in a line meeting 
each of the three given lines; that is, the locus of the point is the hyperboloid 
through the three given lines. It follows that the equations 
Pi, 
Q„ 
Pi, 
Si 
= 0 
p., 
Q„ 
P 2 , 
s 2 
Pa, 
Qä> 
P3, 
Ss 
reduce themselves to a single equation, that of the hyperboloid in question. 
I write for shortness 
(000) = (agh) x 2 + (bhf) y 2 + (cfg) z 2 + (abc) w 2 
+ [(abg) — (cah)] xw 
+ [(bch) - (abf)] yw 
+ [(caf)-(bcg)] zw 
+ [(¥9 ) + W)] yz 
+ [(cgh ) + (afg)] zx 
+ [( a V) + ( h 9 h )] 
viz. (123) will mean (a^g-Ji^) x 2 + etc. where (cq^-A) 
«1, gi, K 
a. 2 , g 2 , h 2 
a 3> ga, K 
etc. denote as usual the determinants 
etc. ; 
then the equations in question are found to be ¿c(123) = 0, y(123) = 0, ,z(123) = 0, 
w(123)=0, reducing themselves to the single equation (123) =0, which is accordingly 
that of the hyperboloid through the three lines ( i ). 
Proceeding now to the above-mentioned problem, we have the point (x, y, z, w), 
and the six lines (eq, b 1} c 1} f 1 , g 1} 7? a ), (a 2 ,...) etc., say the lines 1, 2, 3, 4, 5, 6 : the 
six planes 
P,X + Q 1 Y+ R,Z + &F = 0, etc. 
must be tangents to the same quadricone ; that is, considering the sections by the 
plane W = 0, the six lines 
P 1 X + Q,Y+R 1 Z = 0, etc. 
must be tangents to the same conic, and the condition for this is 
[1 2 3 4 5 6] = 0, 
This equation is given in the paper above referred to, § 54.