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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.

1 Answers 1

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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...

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    If you forget this formula, you can also explicitly do the calculation (i.e. put $x = x^' \cos \theta - y^' \sin \theta$ in your equation). It's not pretty, but it won't take more than three lines.2011-10-10