318 Dr Lachlan, On Algebraic Equations. [May 24, 



These two curves have the point x = z = as a multiple point. 

 In fact G 2m has m branches of the form z 2 + kxy = at this point, 

 and G^ has n branches of a similar form. 



Hence the point x = 0, z = counts for 2mn points of inter- 

 section of the curves. 



Hence the degree of the eliminant 



= 2m x 2n — 2mn = 2mn. 



Ex. 6. The discriminant with respect to x of the equation 



a Q x m + ...+ a r x 2n ~ r + ... + a r 'x r + . . . + a Q ' = 0, 



where a r , a r ' contain y in the rth degree, is of degree 2r 2 with 

 respect to y. 



The discriminant is the same as the eliminant of the 

 equations 



2na x 2n ~i + _ _ _ + (2?i - r ) a r x™- r -^ + ...+ ra^x^ 1 + . . . + a/ = 0, 

 a ^-i + . . . + ra r x 2n - r + ...+ (2w - r) a/x r + ... + 2na ' = 0. 

 Consider then the curves 



CU-i ee 2nu x 2n ~ l + . . . + (2n - r) u r x 2n - r ~ x +... 



+ rurx r - 1 z 2n -^ +...+ u[z™^ = 0, 

 G m = UjX™- 1 + ... + ru r x 2n ~ r + ... 



+ (2n - r) u;x r z 2n - 2r +...+ 2nu 'z m = ; 



where u r , u r ' are homogeneous with respect to y and z of 

 degree r. 



The only common points of G 2n ^ 1 and C^n which lie on the lines 

 x, y, z coincide with x = 0, z = 0. 



C^-! has n — 1 branches of the form z 2 + kxy = passing- 

 through x = 0, z = ; and C m has n similar branches passing 

 through the point. 



Hence the point x = 0, z=0 counts for 2n(n — 1) points of 

 intersection of the curves. 



Hence the degree of the eliminant 



= 2n x (2n - 1) - 2n (n-l)= 2n\ 



(3) Tides on the Equilibrium Theory. By C. Chree, Sc.D., 

 King's College. [Published in Transactions, Vol. XVI. Part II.] 



