# How do you use rotation of axes to identify and sketch the curve of x^2+xy+y^2=1?

Nov 14, 2016

An ellipse
$\frac{1}{2} {\left({x}_{\frac{\pi}{4}}\right)}^{2} + \frac{3}{2} {\left({y}_{\frac{\pi}{4}}\right)}^{2} = 1$

#### Explanation:

This conic can be represented as

${p}^{T} M p = 1$

with $p = {\left(x , y\right)}^{T}$ and $M = \left(\begin{matrix}1 & \frac{1}{2} \\ \frac{1}{2} & 1\end{matrix}\right)$

Making a change of coordinates such that $q = \left({x}_{\theta} , {y}_{\theta}\right)$

$q = {R}_{\theta} p$ with ${R}_{\theta} = \left(\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right)$ we have

$p = {R}_{\theta}^{T} q$ and

${p}^{T} M p = {q}^{T} {R}_{\theta} M {R}_{\theta}^{T} q = {q}^{T} {M}_{\theta} q = 1$

with ${M}_{\theta} = \left(\begin{matrix}1 - \sin \theta \cos \theta & \frac{1}{2} \cos 2 \theta \\ \frac{1}{2} \cos 2 \theta & 1 - \sin \theta \cos \theta\end{matrix}\right)$

Choosing $\theta$ such that $\cos 2 \theta = 0$ or $2 \theta = \frac{\pi}{2}$ or
$\theta = \frac{\pi}{4}$ we have

${M}_{\frac{\pi}{4}} = \left(\begin{matrix}\frac{1}{2} & 0 \\ 0 & \frac{3}{2}\end{matrix}\right)$ and in this new reference frame the conic looks as

$\left({x}_{\frac{\pi}{4}} , {y}_{\frac{\pi}{4}}\right) {M}_{\frac{\pi}{4}} \left(\begin{matrix}{x}_{\frac{\pi}{4}} \\ {y}_{\frac{\pi}{4}}\end{matrix}\right) = \frac{1}{2} {\left({x}_{\frac{\pi}{4}}\right)}^{2} + \frac{3}{2} {\left({y}_{\frac{\pi}{4}}\right)}^{2} = 1$

which is an ellipse.