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
 A: $$8x^2+x\left( 4y^2+4y-40\right) +y^4-11y^2-8y+52=0$$
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 $y \in \{-1,2\}$.
A: For another approach, you can rotate the axes, so that your $xy-$ terms disappear, and you're left with a conic section , i.e., a parabola, ellipse, hyperbola, etc. for which solutions are easy to find. Then go back over your change of axes to get a solution for the original.
The idea is to rotate the plane in such a way to make the $xy-$ terms disappear. The actual transformation is given by $$x=rcos\theta-rsin\theta$$ , and $$y=rsin\theta+rcos\theta$$
You substitute in your equation, and the you set the coefficients of mixed-terms equal to zero. This will give you the angle of rotation. Let me give you a link to a simple example:
http://www.mathamazement.com/Lessons/Pre-Calculus/09_Conic-Sections-and-Analytic-Geometry/rotation-of-axes.html
