498 
SOLUTIONS OF A SMITH’S PRIZE PAPER FOR 1871. 
[537 
We have, from the equation of the conic 
[x + y V(1 + O} 2 = n- (1 + y 2 ), 
that is, 
x + y V(1 + n 2 ) = ± n \/(l + y 2 ), 
but considering the radicals as positive, the sign must be taken so that we have 
simultaneously «=0, y — n. We have therefore 
x + y V(1 + O = n V(1 + y 2 ), 
and similarly 
y + x V(1 + tf) = w V(1 + °^)- 
Then 
n [x V(1 + x 2 ) + y V(1 4- y 2 )} = 2xy + (x 2 + y 2 ) V(1 4- n 2 ) 
= 2xy + V(1 + n 2 ) |?i 2 — 2xy \/(l 4- w 2 )} 
= n 2 {V(l + a 2 ) — 2<ry}, 
which is the first of the relations in question; and 
n 2 {x + V(i + O} {y + V(1 + y-)\ 
= n 2 xy + nx [x + y \/(l + n 2 )] + ny [y + x V(1 + a 2 )} 
+ xy + (x 2 + y 2 ) V(1 4- n 2 ) 4- xy (14- n 2 ) 
= [n + V(1 4- n 2 )} [x 2 + y 2 + 2xy \/(l + n 2 )} 
= [n + a/(1 + w 2 )} n 2 , 
which is the second of the two relations. And the theorem is thus proved. 
[The foregoing is the easiest and most obvious solution, but it is interesting to 
consider the question differently, as follows: 
Write 
we have 
if 
and then 
q_ [ x + V(1 + x 2 )} \y 4- \/(l + y 2 )} 
n + \J{ 1 4- n 2 ) 
Q {^{n 2 41)4- n) = W(1 + x 2 ) + x\ |V(1 4- y 2 ) + y) = A + B, 
Q~ 1 W(n 2 + 1) - n} = {\/(l + x 2 ) - x\ W(1 + y 2 ) -y\ = A-B, 
A = V(1 + a?) V(1 + y 2 ) + xy, 
B = x a/(1 + y 2 ) + y V(1 + æ 2 ); 
AB = x 2 y V(1 + y 2 ) + xy 2 V(1 + x 2 ) 
+ y (l + X 2 ) V(1 4- y 2 ) 4- X (1 + y 2 ) V(1 + O 
= X V(1 + X 2 ) + y V(1 + y 2 ) + 2xyB,