<s>[INST] How solve equation:$y^4+4y^2x-11y^2+4xy-8y+8x^2-40x+52=0$ Let $x,y\in \mathbb{R}$, solve this follow equation:$$y^4+4y^2x-11y^2+4xy-8y+8x^2-40x+52=0$$
My try: Since 
$$8x^2+4x(y^2+y-10)+y^4-11y^2-8y+52=0$$
then I can't. Maybe this problem have other nice methods.Thank you
 [/INST] $$8x^2+x\left( 4y^2+4y-40\right) +y^4-11y^2-8y+52=0$$
This is quadratic in $x$, thus
$$x_{1,2}=\frac{-4y^2-4y+40\pm \sqrt{\left( 4y^2+4y-40\right)^2-32\left( y^4-11y^2-8y+52\right)}}{16}$$
Now, use
$$\left( 4y^2+4y-40\right)^2-32\left( y^4-11y^2-8y+52\right)=-16(y-2)^2(y+1)^2$$
to show that the equation has solutions if and only if $y \in \{-1,2\}$.
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