What subject does this fall under (differential curves maybe?) I found a question where I don't even know where it comes from (except, vaguely, a Calc 2 class--but in my past Calc 2 class I never saw anything like this):

Find the angle of rotation needed to eliminate $xy$ from the equation $2x^2 +2\sqrt{3}xy+4y^2 +8x +8\sqrt{3}y=50$.

Can anyone tell me the basic method of doing this, or at least what topic this kind of problem falls under?  Is this some kind of change of coordinates?  Does that mean that you need to find two new parameters in an orthonormal basis such that the equation becomes ... linear?  Or just a polynomial?
 A: Say that $Q\colon \mathbb R^2\rightarrow \mathbb R$ is the function that maps $(x,y)$ to $2x^2 +2\sqrt{3}xy+4y^2 +8x +8\sqrt{3}y$. Note that the set of points $(x,y)$ so that $Q(x,y)=50$ is an ellipse (picture).
Let $e_1$ be $(1,0)$ and $e_2$ be $(0,1)$, the vectors of the standard basis for $\mathbb R^2$.
Then we have:
$\begin{align*}
Q(xe_1 + ye_2) = 2x^2 +2\sqrt{3}xy+4y^2 +8x +8\sqrt{3}y &= 2\left[x^2 + \sqrt{3}xy + 2y^2\right]+8x+8\sqrt 3y\\
 &= 2(x+\dfrac{\sqrt 3}{2}y)^2 + \dfrac{5}{2}y^2 + 8x + 8\sqrt{3}y.
\end{align*}$
Now we want to find a new basis $e_1', e_2'$ so that the expression of $Q$ in this basis
does not contain $xy$. Here, it suffices to change $x+\frac{\sqrt 3}{2}y$ to $x'$:
let $e_1' = e_1$ and $e_2' = -\frac{\sqrt 3}{2}e_1 + \frac{1}{2}e_2$.
Now, we have $Q(xe_1' + ye_2') = Q((x-\frac{\sqrt{3}}{2}y)e_1 + ye_2) = 2x^2 + \dfrac{5}{2}y^2 + 8x + 4\sqrt 3y$.
Basically, we tilted the $y$-axis so that it is parallel with one of the axis of the ellipse. Now, it simply remains to compute the angle between $e_2'$ and $e_2$, which you can do with a simple dot product ($e_2$ and $e_2'$ have norm $1$).
