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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.

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up vote 4 down vote accepted

There is a nice formula for the rotation angle needed to remove the cross-term of a general conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$:


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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And how exactly does one obtain that formula? – Huy Oct 10 '11 at 16:52
Then you should see this. Proving the formula for $\tan\,2\theta$ ought to have been proven in your course... – J. M. Oct 10 '11 at 16:56
I am very sure that we have not even seen any formula - be it for $\tan 2 \theta$ or for $\tan \theta$. This exercise was given in Physics I and all we've learnt there is what the rotation matrix does. – Huy Oct 10 '11 at 17:01
A sketch: you make the substitutions $x=x^\prime \cos\,\theta-y^\prime \sin\,\theta$ and $y=x^\prime \sin\,\theta+y^\prime \cos\,\theta$. Collect terms and note the coefficient of $x^\prime y^\prime$. Equate it to $0$... – J. M. Oct 10 '11 at 17:09
For the general $Ax^2 + Bxy + Cy^2 = D$, I end up with $\sin(2 \varphi) (C-A) + \cos(2 \varphi) B = 0$. For this equation, wolframalpha gives terrible solutions. What did I do wrong if anything? Or will it simplify when I solve it for $\tan \varphi$? – Huy Oct 10 '11 at 17:32

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