0
$\begingroup$

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.

|cite|improve this question
$\endgroup$
4
$\begingroup$

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...

|cite|improve this answer
$\endgroup$
  • $\begingroup$ And how exactly does one obtain that formula? $\endgroup$ – Huy Oct 10 '11 at 16:52
  • $\begingroup$ Then you should see this. Proving the formula for $\tan\,2\theta$ ought to have been proven in your course... $\endgroup$ – J. M. is a poor mathematician Oct 10 '11 at 16:56
  • $\begingroup$ 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. $\endgroup$ – Huy Oct 10 '11 at 17:01
  • $\begingroup$ 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 $0$... $\endgroup$ – J. M. is a poor mathematician Oct 10 '11 at 17:09
  • $\begingroup$ 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 $\tan \varphi$? $\endgroup$ – Huy Oct 10 '11 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.