157] 
ON THE TANGENTIAL OF A CUBIC. 
559 
where the second line is in fact equal to zero, on account of the first factor, which 
vanishes. And Qf, 15 denote respectively quadric and cubic functions of (;y, z), which 
are to be determined so as to make the right-hand side divisible by x 2 ; the resulting 
value of £ may be modified by the adjunction of the evanescent term 
(a^+hy+j£)(a, b, c, f, g, h, i, j, k, IJoc, y, zf, 
where a, h, j are arbitrary coefficients ; but as it is not obvious how these coefficients 
should be determined in order to present the result in the most simple form, I have 
given the result in the form in which it was obtained without the adjunction of any 
such term. 
Write for shortness, 
P = (k,l 1y,z), 
Q = Q>, f Î $y, zf, 
R = (9’ $y. z ) » 
£ = (/> i c Jy, zf, 
B = ( h > j «) » 
C=(k, l, g ffy, zf, 
D = (6, /, i, cfty, zf, 
(h, b, i, f, l, k \x, y, zf = (h, P, Q $>, If, 
(j, f, c, i, g, l Jx, y, zf = (j, R, S $>, l) 2 , 
(a, b, c, f, g, h, i, j, k, Iffx, y, zf-(a, B, C, D$>, l) 3 . 
<Sx + 15 = (©, IB fix, 1), 
and then for greater convenience writing (h, 2P, Qf[x, l) 2 , &c. for (h, P, Qf[x, l) 2 , &c., 
and omitting the (x, If, &c. and the arrow-heads, or representing the functions simply 
by (h, 2P, Q), &c., we have 
x 2 !j = b (j, 2R, S ) 3 
-Sf(j,2R,S f.(h,2P,Q) 
+ 3i{j, 2R, S ) .(h, 2P, Qf 
- c . (h , 2P, Qf 
- (a, SB, SG, D) . (Qf, 15 ), 
which can be developed in terms of the quantities which enter into it. The con 
ditions, in order that the coefficients of x, x° may vanish, are thus seen to be 
1)15 = bS 3 - SfS 2 Q + 3iSQ 2 - cQ 3 , 
DQf - 3(715 = b (6RS 2 ) - 3/(2S 2 P + 4RSQ) + 3» (2RQ 2 + 48PQ) - c (6PQ 2 ), 
and from these we obtain 
<K 
bck — 3 
big + 6 
beg + 3 
bil + 6 
cfk — 6 
cfl -6 
fik + 3 
/V-6 
fgi -3 
fH -6 
i 2 k + 6 
m +6