ON POLYZOMAL CURVES. 
567 
414] 
(r — s') (r — s) points of intersection of the curves © = 0, 4> = 0, each of these points is 
a node on the Jacobian, and hence that the Jacobian must be of the form 
J(Z© + L<&, m© + M®, «© + №) = A© 2 + 2i?©4> + C4 32 = 0, 
where obviously the degrees of A, B, C must be r + 2s — 3, r + s + s' — 3, r + 2s — 3 
respectively. In the particular case where s' = 0, that is where l, m, n are constants, 
we have A = 0; the Jacobian curve then contains as a factor (4> = 0), and throwing 
this out, the curve is B© + (74? = 0, viz., this is a curve of the order 2r + s — 3 
passing through each of r (r — s) points of intersection of the curves © = 0, 4> = 0. 
In particular, if r = 2, s = 1, that is, if the curves are the conics © + L<& = 0, 
© + _M4> = 0, © + ATfi = 0, passing through the two points of intersection of the conic 
© = 0 by the line 4> = 0, then the Jacobian is a conic passing through these same 
two points, viz., its equation is of the form © + 04? = 0. This intersects any one of 
the given conics, say © + A4> = 0 in the points © = 0, 4> = 0, and in two other points 
© + i!4> = 0, il — L = 0; at each of the last-mentioned points, the tangents to the two 
curves, and the lines drawn to the two points © = 0, = 0, form a harmonic pencil. 
Although this is, in fact, the known theorem that the Jacobian of three circles 
is their orthotomic circle, yet it is, I think, worth while to give a demonstration of 
the theorem as above stated in reference to the conics through two given points. 
Taking (z = 0, x — 0), (z= 0, y = 0) for the two given points © = 0, <E> = 0, the 
general equation of a conic through the two points is a quadric equation containing 
terms in z 2 , zx, zy, xy; taking any two such conics 
cz 1 + 2fyz + 2 gzx + 2 hxy — 0, 
Gz 1 + 2 Fyz + 2 Gzx + 2Hxy — 0, 
these intersect in the two points (x = 0, 2 = 0), {y = 0, 0 = 0) and in two other points; 
let (x, y, z) be the coordinates of either of the last-mentioned points, and take (X, Y } Z) 
as current coordinates, the equations of the lines to the fixed points and of the two 
tangents are 
Xz-Zx = 0, Yz-Zy =0, 
(hy + gz) (Xz — Zx) + (hx +fz ){Yz — Zy) = 0, 
(Hy + Gz) (Xz - Zx) + (Hx + Fz) (Yz - Zy) = 0, 
whence the condition for the harmonic relation is 
(hy + gz) (Hx + Fz) 4- (hx + fz) (Hy + Gz) = 0, 
that is 
(fG + gF) z- + (IiF + fH) yz + (gH + hG) zx + 2hHxy = 0, 
but from the equations of the two conics multiplying by \H, \h and adding, we have 
| (cH + hC) z~ + (hF + fH) yz + (gH +hG)zx + 2hHxy = 0 ; 
viz., the condition is thus reduced to 
cH+ hC—2 (fG+gF) = 0,