Eliminate xy term of conic

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.

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.
• Would $\theta$ be $pi/6$ or $pi/3$? I've tried both with no luck. May 16, 2014 at 3:35
• You would get $\theta = \pi/6$, from which the desired transformation is $$x = \tfrac{1}{2}(\sqrt{3}x'+y'), \quad y = \tfrac{1}{2}(\sqrt{3}y' - x').$$ Substituting this directly into the equation gives $$\tfrac{3}{2}x^2 + \tfrac{15}{2}y^2 - 11 = 0.$$ Notice that the other choice of angle, $\theta = -\pi/3$, would give $$\tfrac{15}{2}x^2 + \tfrac{3}{2}y^2 - 11 = 0,$$ which is also admissible--it simply depends on if you want the longer axis to be parallel to the $x$-axis or $y$-axis. May 16, 2014 at 3:53
• Thanks for explaining that. The final answer I got was $(3/22)x^2 + (15/22)y^2 = 1$. This didn't seem correct since there were numbers on the top of the fraction. Am I done, or do I need to go further? May 16, 2014 at 3:58
• I don't know what else you want to do with this result. If all you want is to rotate the ellipse to eliminate the $xy$ term, then you're done. May 16, 2014 at 4:02
Use the formulæ for rotating through an angle $\theta$, where $\tan{2\theta}={{3\sqrt 3}\over{3-6}}$.