$\displaystyle x^2 + 2\sqrt{3}xy + 3y^2 + 8\sqrt{3}x - 8y + 32 = 0$

A) Use the Discriminant test to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola.

B) Find a suitable rotation of axis

C) Find an equation for the graph in a $\displaystyle x'y'$ - plane

D) Sketch the graph over top both coordinate systems with your angle of rotation displayed.

My Work:

A) $\displaystyle B^2 - 4AC \rightarrow (2\sqrt{3})^2 - 4(1)(3) = 0$ So, The equation is a parabola.

B) Here's where I get somewhat confused, so bear with my work.

$\displaystyle tan 2\alpha = \frac{B}{A-C} \rightarrow \frac{2 \sqrt{3}}{1-3} \rightarrow \frac{\sqrt{3}}{-1}$

$\displaystyle 2 \alpha = tan^{-1} \left(\frac{-\pi}{3}\right)$

$\displaystyle \alpha = \frac{-\pi}{6}$

From here on I'm not really quite sure what to do. The notes I have on this don't really give me enough help to continue on. I know later on though I have to utilize these:

$\displaystyle A'=Acos^2 \alpha + Bcos\alpha sin\alpha + Csin^2 \alpha$

$\displaystyle B'=Bcos2\alpha + (C-A)sin2 \alpha$

$\displaystyle C'=Asin^2 \alpha - Bcos\alpha sin\alpha + Ccos^2 \alpha$

$\displaystyle D'=Dcos\alpha + Esin\alpha$

$\displaystyle E'=-Dsin\alpha + Ecos\alpha$

$\displaystyle F'=F$