INVOLUTIVE 1-1 QUADRIC TRANSFORMATION IN A PLANE. 255 



The transformed equations are 



(1)' a<A 1 £+B l9 +C 1 )+y(A 8 £+ B 2 ^ + C 2 ) + A 3 f + B aV + C S = 

 (2)' x(a 1 g+ etc. )+ etc. =0. 



Case 1. If 



(2) s (2)' 



there must exist the following equations in the coefficients of ( 1 ) — 



A 1 = A 1 , A, = B V A 3 = C 1; 

 B 1 = A 2 , B 2 — B 2 , B 3 = C.„ 



C 1= A3, c 2 = b 3 , c 3 =c 3 , 



so that (1) may be written as 



I. K x £+ ^m + C 3 + G& + + C 2 (y + r}) + \{£y + x v ) = 0. 

 But if A-! = B 1 = C± = 0, we may reverse signs throughout and deduce 



II. G 1 (x±£) + G 2 (y±ti) + B x (^±^) = 0. 



Similar results hold for the second equation. (I. is practically the only case dis- 

 cussed by Czuber.) 



Case 2. Nothing new is gained by supposing 



(1)' S ±(2); (2) s ±(l). 

 For the solutions of (1) = and (2) = are those of 



(l) + (2) = 0; (l)-(2) = 0, 

 and the transformation transforms these latter equations into themselves, so that there 

 is a reduction to the preceding case. 



Case 3. More generally, no new result is obtained by supposing 



(l)' = k(l) + l(2) (i.) 

 (2)' s m(l)+ % (2) (ii.); 



for, since the repetition of the transformation gives again (l) = and (2) = 0, there 

 would result 



(1) = k{k{l) + l(2)} + l{7n(l) + n(2)} 



(2) ^ m{k(l) + l(2)} +n\m(l) + n(2)} 



i.e., 



(Y) = {W+lm)(l)+l{k+n)(2) 

 (2) = m(k + n)(D + (lm+n 2 )(2), 



giving 



k 2 + lm = l\ 

 l(h+n) = ..... 



lm + n 2 =l ' 



m(k+/i) = y 

 If ?=f=0, -m=h0, these equations reduce to 



