ON POLYZOMAL CURVES. 
527 
414] 
or, what is the same thing, 
£77 (pi; + qi1 + ez) + z 2 («£ + br} + cz) = 0, 
viz. (£, r], z) being any coordinates whatever, this is the general equation of a cubic 
passing through the points (£ = 0, z— 0), (77 = 0, 2 = 0), and at these points touched by 
the lines £ = 0, 77 = 0 respectively. And if (f, 77, z= 1) be circular coordinates, then we 
have the general equation of a circular cubic having the lines £ = 0, 77 = 0 for its 
asymptotes, or say the point £ = 0, 7; = 0 for its centre; the equation of the remaining 
asymptote is evidently pi; + qrj + ez = 0; to make the curve real we must have (p, q) 
and (a, b) conjugate imaginaries, e and c real. 
143. Taking in any case the points /, J to be the points £ = 0, z — 0 and 7; = 0, 
z = 0 respectively, for the equation of a tangent from I write pi; — 6z; then we have 
07) (6z + qt) + ez) + z (adz + bprj + cpz) = 0, 
that is 
z 2 (a0 + cp) + 7)z (6 2 + ed + bp) + t; 2 . q0 = 0, 
and the line will be a tangent if only 
(O' 2 + ed + bp) 2 — 4>q0 (a6 + cp) = 0, 
that is, the four tangents from I are the lines pi; = 0z, where 6 is any root of this 
equation; similarly the four tangents from J are the lines qTj = <f>z, where <£ is any 
root of the equation 
((f) 2 + e(f> + aq) 2 — 4p<j> (bcf) + cq) = 0. 
Writing the two equations under the forms 
( 6 ^ 
u > 
Se, 
e 2 + 2bp - 4aq , £0, l) 4 = 0, 
3 ebp — 6cpq, 
< 6& 2 i> 2 , j 
'6, 
Se. 
- e 2 + 2aq — 46p , <£(f>, l) 4 = 0, 
Seaq — Qcpq, 
k 6a?q 2 , J 
the equations have the same invariants; viz., for the first equation the invariants are 
easily found to be 
I = 3 (e 2 — 46p — 4ag) 2 +72 (ce — 2ab) pq, 
J = — (e 2 — 46p — 4aq) 3 — 36 (ce — 2ab) pq (e 2 — 46p — 4ag) — 216 c 2 p 2 q 2 , 
and then by symmetry the other equation has the same invariants. The absolute 
invariant I 3 J 2 has therefore the same value in the two equations ; that is, the 
equations are linearly transformable the one into the other, which is the before- 
mentioned theorem that the two pencils are homographie. 
144. 
a 
P = k’ q 
The two equations will be satisfied by 0 = (f), if only bp = aq\ that 
= \; putting for convenience j in place of e, the equation of the curve is 
Ic № 
£77 (ai; + 677 + ez) + Jcz 2 (a% + 677 + cz) = 0. 
is, if 
then