Rotating a conic equation

Question

Identify the type of conic, transform the equation in x and y into equation in X and Y (without an XY term) by rotating the x- and y- axes by $θ=30^\circ$ to arrive at the new X- and Y- axes and graph the resulting equation showing both sets of axes.

$7x^2 + 4\sqrt{3}xy + 3y^2 - 9 = 0$

EDIT: There was a typo. It is fixed

Umm. That's not really a conic equation as it stands. Are you sure the equation is supposed to have that $\sqrt{x}$ term? — sharding4, Jun 10 '17 at 16:35
You can look at the discriminant = $(4\sqrt{3})^2-4\cdot 7\cdot 3 = -36$ to determine that you have an ellipse. What are you having problems with? — sharding4, Jun 10 '17 at 16:42
You want $x = X \cos 30^{\circ} - Y \sin 30^{\circ}$ and $y = X \sin 30^{\circ} +Y \cos 30^{\circ}$ — sharding4, Jun 10 '17 at 16:49

score 1 · Answer 1 · answered Jun 10 '17 at 16:50

Let $\displaystyle \begin{pmatrix}x \\ y\end{pmatrix}=\begin{pmatrix}\cos30^\circ & -\sin30^\circ\\ \sin30^\circ & \cos30^\circ\end{pmatrix}\begin{pmatrix}X \\ Y\end{pmatrix}=\frac{1}{2}\begin{pmatrix}\sqrt{3} & -1\\ 1 & \sqrt{3}\end{pmatrix}\begin{pmatrix}X \\ Y\end{pmatrix}$.

Then do the substitution.

score 0 · Answer 2 · answered Jun 10 '17 at 17:44

This can be done very nicely in the complex plane. We first convert to polar form with $x=r\cos\theta,\,y=r\sin\theta$ so that

$$7x^2 + 4\sqrt{3}xy + 3y^2 - 9 = 0\to r^2(7\cos^2\theta+4\sqrt{3}\cos\theta\sin\theta+3\sin^2\theta)=9\\ r=\sqrt{\frac{9}{7\cos^2\theta+4\sqrt{3}\cos\theta\sin\theta+3\sin^2\theta}}$$

Then we can say

$$z=re^{i\theta},\quad \theta\in[0,2\pi]$$

And finally, any rotation, say $z_{\theta_0}$, is given by $z_{\theta_0}=ze^{i\theta_0}$.

Rotating a conic equation

2 Answers2