Graph $\displaystyle x^2-2\sqrt{3}xy+3y^2-12\sqrt{3}x-12y$

I seem to be getting lost in the tedious amount of algebra involved, please find my mistakes:

$\displaystyle \text{Angle=}cot2\theta=\frac{1}{\sqrt{3}}=\frac{6 0^\circ}{2}=30^\circ$

$\displaystyle x=x'cos30^\circ-y'sin30^\circ=\frac{x'\sqrt3-y'}{2}$

$\displaystyle y=x'sin30^\circ+y'cos30^\circ=\frac{x'\sqrt3+y'}{2 }$

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$\displaystyle x^2-2\sqrt{3}xy+3y^2-12\sqrt{3}x-12y$

$\displaystyle (\frac{x'\sqrt3-y'}{2})^2-2\sqrt3(\frac{x'\sqrt3-y'}{2})(\frac{x'\sqrt3+y'}{2})+3(\frac{x'\sqrt3+y' }{2})^2-12\sqrt3(\frac{x'\sqrt3-y'}{2})-12(\frac{x'\sqrt3+y'}{2})$

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$\displaystyle (\frac{3(x')^2-2\sqrt3x'y'+(y')^2)}{4}-2\sqrt3(\frac{3(x')^2-(y')^2}{4}+3(\frac{3(x')^2+2\sqrt3x'y'+(y')^2}{4}-12\sqrt3(\frac{x'\sqrt3-y'}{2})-12(\frac{\sqrt3x'+y'}{2})=0$

The y^2 term ends up giving me another x'y' term that I find to be irreducible, which is nonsense because the equation is a fairly simple parabola.

I would continue but I am certain the problem stems from this area or earlier.