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.
There is a nice formula for the rotation angle needed to remove the cross-term of a general conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$:
$$\tan\,\theta=\frac{B}{A-C+(\mathrm{sign}\;B)\sqrt{B^2+(A-C)^2}}$$
In this case, $\tan\,\theta=\dfrac1{\sqrt 3}$; this yields the rotation matrix $$\begin{pmatrix}\frac{\sqrt 3}{2}&-\frac12\\\frac12&\frac{\sqrt 3}{2}\end{pmatrix}$$
I'll let you do the rest...