84 
ON CERTAIN OCTIC SURFACES. 
[654 
systems has, in reference to any pair of nodal conics, a homographic property as 
follows; viz. considering for example the two conics in the planes z — 0 and to = 0 
respectively, if a line meets these conics in the points P and Q respectively, and 
through these points respectively and the line x = 0, y — 0 we draw planes, then the 
system of the P planes and the system of the Q planes correspond homographically 
to each other, the coincident planes of the two systems being the planes x = 0 and 
y = 0 respectively. 
Conversely, if through the line (x = 0, y = 0) we draw the two homographically 
related planes meeting the two conics in the points P and Q respectively, then, for 
a proper value (determined by a quadratic equation) of the constant ¿(=@4-^) which 
determines the homographic relation, the line PQ will be a line meeting each of the 
four conics, and will belong to one or other of the above-mentioned two systems, 
as k is equal to one or the other of the two roots of the quadratic equation. The 
scroll generated by the lines meeting each of the four conics, or what is the same 
thing, any three of these conics, is primd facie a scroll of the order 16; but by 
what precedes, it appears that this scroll breaks up into two scrolls, which will be 
each of the order 8. Moreover, each scroll has the four conics for nodal curves; and 
since the equation U = 0 is the general equation of an octic surface having the 
four conics for nodal curves, it follows, that the equation of the scroll is derived 
from that of the octic surface U= 0, by assigning a proper value to the indeterminate 
eoefficient O; so 'that there are in fact two values of fi, for each of which the 
surface U = 0 becomes a scroll. 
To sustain the foregoing conclusions, take x — 0'y, x = 0y for the equations of the 
two planes through the line (x = 0, y = 0), which meet the 2-conic and w-conic in the 
points P and Q respectively; then the equations of the line PQ are 
f(/0 2 + g) (x — 6'y) + V(- h) (0' — 0)2=0, 
— f(b0' 2 — a) (x — 6y ) 2 + V(— h) (6' — 6) io = 0, 
or, writing therein 0' = kd, the equations are 
f(f0 2 + g)(x — k6y) + V(- h) (k-1)02 = 0, 
— f(bk 2 0 2 — a) (x — 0y) + \/(— h) (k — 1) div = 0. 
To find where the line in question meets the plane y = 0, we have 
V(/# 2 + g) x + h) {k — 1) 02 = 0, 
— f(bk-0 2 — a) x + h) (k — 1) 0to = 0, 
and thence 
(f0 2 + g) a? + h {k — l) 2 0 2 2 2 = 0, 
(bk 2 0 2 — a) x 2 + h (k — l) 2 0Ho 2 = 0, 
or multiplying a, g and adding 
(a/+ bgk 2 ) x 1 + h (k — l) 2 {az 2 + gw 1 ) = 0, 
af-1- bgk 2 + cli {k— 1 ) 2 = 0, 
or assuming