140 Proceedings of Royal Society of Edinburgh. [sess. 
of eliminating y from tlie two equations. It will be found tbat 
the cubic in k is identical witb the reducing cubic of the biquadratic 
in x . 
As an illustration, consider the so-called homogeneous system 
U = ax 2 + bxy + cy 2 + d = 0 i 
V = a'x 2 + b’xy + c'y 2 + d' = 0 j 
This system may be solved by putting y\x = and obtaining a 
quadratic in v. This method of solution is an algebraical interpreta- 
tion of the fact that the points of intersection of two concentric 
conics are the vertices of a parallelogram. The above system may 
also be solved by making use of the fact that - d/d' is one value of 
Jc for which U + kV may be resolved into factors ; and this solution 
is practically the same as that obtained by putting y/x = v. The 
question then naturally arises, what are the other two values of k 
which make U + kY resolvable into factors ? These values are the 
values for which ( ax 2 + bxy -f cy 2 ) + k(a'x 2 + b’xy + c'y 2 ) is a complete 
square ; and this suggests another method of solution. 
By way of further illustration, a method of obtaining the reducing 
cubic of the biquadratic +px 2 + qx + r = 0, may be noticed. 
The resultant in x of the two equations 
x 2 -y + p 2 = 0 ) . , o 
o o / „ ^ V is Xr + px 2 + qx + r = 0 . 
y 2 + qx + r-p 2 / 4 = 0 j 
Now the equation to determine k so that 
(x 2 — y +pl 2) + Jc(y 2 + qx + r — p 2 /4) 
may be resolvable into factors is 
k s - 2 'pk 2 - (4 r ~p 2 )k + q 2 = 0 , 
which is the reducing cubic of 
a? 4 +px 2 + qx + r = 0 . 
[The resultant in y is 
?/ 4 + (2 r - p 2 /2)y 2 - q 2 y + (j» 4 /16 + r 2 -p 2 r/2 +pq 2 /2) = 0 , 
the reducing cubic of which is 
k 3 - ( 4 r -p 2 )k 2 - 2p<fk + f = 0 , 
