316 Dr Lachlan, On the degree of the [May 24, 



Introducing £, 77 to make the equations homogeneous, we may 

 write them in the form 



a=-%^+-^+...+-%-j-i=o J 



Ojrj + a 2 rj + a n 7] + V 



a x ?7 + a a 2 7) + a n i) + 



G and C are curves of the nth degree, having n common points 

 on the line £ = 0, viz. 



of) + = 0, a 2 rj + = 0, &c, 



and (w — 1) branches passing through the point 77 = = 0, and no 

 branch of one curve touches a branch of the other. 



Hence the order of the eliminant is 



n 2 — n — (n — l) 2 = n — 1. 



This result may also be inferred from § 2, for if £ be 

 eliminated, the result is obviously of degree (n— 1) with respect 

 to t] and 0. 



Ex. 2. The discriminant with respect to of the equation 



xAJ"\ wn d J 'il. -t 



1 I ? 1- -1 — = 1 



Oi + a 2 +<T a n + ' 



is of the degree 2 (n - 1). 



We have to find the degree of the eliminant of 



+ S - + ... + S " =1, 



a-w + a 2 r) + a n r\ + 



'+ /„ _ . /i\a + • • • + /„ _ , a\i - u > 



(CM7 + 0) 2 (a 2 T7 + <9) 2 " (a n 7? + 0f 



while the latter equation represents 2 (?i — 1) lines through the 

 point r) = = Q, the degree of the eliminant must be 2(n— 1) 

 (§ 2). 



22*. 3. If be eliminated from the equations 



ft* ft* ft* 



\Aj-\ tAjn ww _. 



(^ + 0)* (a 2 +0)P '■* (a n + 0)* 



-"•1^1 . -"-2^2 . . -"-W^W 1 



+ /_ . a\n + ■ ■ ' + TZ. r~a^ n = L > 



the degree of the eliminant is np — q (p > q) 



