For the following problem, I am trying to eliminate xy, and I've tried numerous times to solve if with no luck. I need to find the general form of the equation rotated to eliminate the xy term.
$ 3x^2+3\sqrt{3}xy+6y^2-11=0 $
When I've tried to solve it, I got the general form to be the following:
$ (15/2)x^2 + (3/2)y^2 - 11 = 0 $
This didn't seem right. When I've tried solving it, I've been able to get A' to be 3/2, 15/2 and 9/2, and vice versa for C'.
I only need to see how to get A' correctly, and then I should be able to get C' and finish the problem.
$\endgroup$2 Answers
$\begingroup$The idea is that a suitable rotation of the form $$\begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{bmatrix}\begin{bmatrix}x' \\ y' \end{bmatrix}$$ will eliminate the $xy$ term. Substituting and solving for the condition $x'y' = 0$ gives $\tan 2\theta = \frac{B}{C-A}$. Therefore, evaluating the relationship for a specific set of coefficients gives you the desired angle $\theta$, from which you can calculate the transformation required.
$\endgroup$ 9 $\begingroup$Use the formulæ for rotating through an angle $\theta$, where $\tan{2\theta}={{3\sqrt 3}\over{3-6}}$.
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