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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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    And how exactly does one obtain that formula?2011-10-10
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    Then you should see [this](http://math.stackexchange.com/questions/70069). Proving the formula for $\tan\,2\theta$ ought to have been proven in your course...2011-10-10
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    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.2011-10-10
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    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 $0$...2011-10-10
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    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 $\tan \varphi$?2011-10-10
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    Good! So $\tan\,2\varphi$ is equal to what? You can then derive the expression for $\tan\,\varphi$ using the double angle formula and the quadratic equation...2011-10-10
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    Then, $\tan 2 \varphi = \frac{B}{A-C}.$ Now, is there any reason why I shouldn't simply take the arctangent on both sides and then divide by 2?2011-10-10
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    You certainly could; I'm just not fond of computing angles from arctangents with the subsequent plugging of these angles into sines and cosines... when the whole thing can be done entirely algebraically. I made a similar remark in the thread I linked to.2011-10-10
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    I see. Okay, now, everything is clear. Thank you very much for the help.2011-10-10
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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