Question

If $(x,y)$ satisfies the simultaneous equations \begin{align*} 3xy - 4x^2 - 36y + 48x &= 0, \\ x^2 - 2y^2 &= 16, \end{align*}where $x$ and $y$ may be complex numbers, determine all possible values of $y^2$.

Answer 1

$3xy-4x^2-36y+48x=3y(x-12)-4x(x-12)=(3y-4x)(x-12)=0$


So either $3y=4x$, or $x=12$. In the first case, we find


$x^2-2y^2=16\implies x^2-2\left(\dfrac{4x}3\right)^2=16\implies x^2=-\dfrac{144}{23}$


from which it follows that


$-\dfrac{144}{23}-2y^2=16\implies y^2=-\dfrac{256}{23}$

Alternatively, if $x=12$, then

$12^2-2y^2=16\implies y^2=64$