100 PROFESSOR CAYLEY ON POLYZOMAL CURVES. 



the order s', L = 0, M = 0, N = each of the order s) which have in common 

 the (r — s') (r — s) points of intersection of the curves = 0, <£ = 0, each of these 

 points is a node on the Jacobian, and hence that the Jacobian must be of the 

 form 



J(l& + L$, m© + M<P, n® + iVfe) = A& 2 + 2B&P + C$ 2 = , 



where obviously the degrees of A, B, C must be r + 2s'— 3, r+8+8*— 3, r + 2s— 3 

 respectively. In the particular case where s' = 0, that is where /, m, n are con- 

 stants, we have A=0; the Jacobian curve then contains as a factor (<|> = 0), and 

 throwing this out, the curve is Bq + C$ = 0, viz., this is a curve of the order 

 2r + s — 3 passing through each of r(v — s) points of intersection of the curves 

 9 = 0, $ = 0. 



In particular, if r = 2, a — I, that is, if the curves are the conies 

 + Z$> = 0, + M$> — 0, + N$ — 0, passing through the two points of 

 intersection of the conic = by the line 4> = 0, then the Jacobian is a conic 

 passing through these same two points, viz., its equation is of the form 

 + Q3> = 0. This intersects any one of the given conies, say + L$> = in the 

 points = 0, <& = 0, and in two other points + Q$> = 0, Q — L = ; at each 

 of the last-mentioned points, the tangents to the two curves, and the lines drawn 

 to the two points = 0, 4> = 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 conies through two given points. 



Taking (z = 0, x — 0) (z = 0, y — 0) for the two given points = 0, $ = 0, 

 the general equation of a conic through the two points is a quadric equation con- 

 taining terms in z 2 , zx, zy, xy ; taking any two such conies 



cz> + 2fyz + 2gzx + 2hxy = , 

 Cz 2 + 2Fyz + 2Gzx + 2Hxy = , 



these intersect in the two points (x = 0, z — 0), (y — 0, z = 0) and in two other 

 points ; let (x, y, z) be co-ordinates of either of the last-mentioned points, and 

 take (X, Y, Z) as current co-ordinates, the equations of the lines to the fixed 

 points and of the two tangents are 



Xz - Zn = , Yz - Zy = , 

 (hy + gz ) {Xz - Zx) + (hx + fa ) ( Yz - Zy) = , 



{Hy + Gz) (Xz - Zx) + (Hx + Fz) (Yz - Zy) = , 



whence the condition for the harmonic relation is 



(hy + gz)(Ex + Fz) + (hx +fz) (Hy + Gz) = , 

 that is 



(fG + gF)# + (hF+/H)yz + (gH + hG)zx + 2hHxy = , 



