I'm currently trying to solve the following problem:
Let $L$ be the set of points of $\mathbb{R}^2$ that satisfy the condition $f(x,y) = 7x^2-6 \sqrt{3} xy + 13y^2 = 16$. It is possible to apply a rotation of the axes such that if you call (x',y') the coordinates with respect to the new reference system, the condition takes the form h(x',y')=1, but now there is no x'y' term in h(x',y'). Find h(x',y').
Now my idea was to use a rotation matrix to rotate the coordinate system, that is to say I wrote
$\mathbf{v} = \begin{pmatrix} x\\ \frac{1}{13} (3 \sqrt{3} \pm 4 \sqrt{13-4x^2}) \end{pmatrix}$
and then \mathbf{v}' = R^{-1}(\varphi)\cdot \mathbf{v}, where
$R(\varphi) = \begin{pmatrix} \cos \varphi & - \sin \varphi\\ \sin \varphi & \cos \varphi \end{pmatrix}.$
Now the problem is that I have a vector of the form \mathbf{v}' = (f(x), g(x))^T and I don't know how to transform it such that it is in the form \mathbf{v}' = (x, y(x))^T - if it was in the latter form, I could easily write down h(x',y').
So my question is: How can I transform the vector I have to the form I want it to be? Is this even the correct approach or is there a far easier way to solve this exercise which I have not seen?
Thank you very much for any answer.