How do you graph the conic 13x^2+6sqrt3xy+7y^2-16=0 by first rotations the axis and eliminating the xy term?

Aug 29, 2016

$16 {X}^{2} + 4 {Y}^{2} - 16 = 0$ An ellipse.

Explanation:

$13 {x}^{2} + 6 \sqrt{3} x y + 7 {y}^{2} - 16 = 0$ is equivalent to

${\left(\begin{matrix}x \\ y\end{matrix}\right)}^{T} \left(\begin{matrix}13 & 3 \sqrt{3} \\ 3 \sqrt{3} & 7\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) - 16 = 0$

Introducing a change of variables

$\left(\begin{matrix}X \\ Y\end{matrix}\right) = \left(\begin{matrix}\cos \theta & - S \int h \eta \\ S \int h \eta & C o s \theta\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right)$

we get at

((X),(Y))^T((10 + 3 Cos2theta - 3 sqrt[3] Sin2theta, 3 (sqrt[3] Cos2theta + Sin2theta)),(3 (sqrt[3] Cos2theta + Sin2theta), 10 - 3 Cos2theta + 3 sqrt[3] Sin2theta))((X),(Y))-16=0

Choosing $\theta$ such that

$3 \left(\sqrt{3} C o s 2 \theta + S \in 2 \theta\right) = 0$

we have $\theta = \frac{\pi}{3}$ or $\theta = - \frac{\pi}{6}$

$16 {X}^{2} + 4 {Y}^{2} - 16 = 0$

or

$4 {X}^{2} + 16 {Y}^{2} - 16 = 0$

In both cases the quadratic is characterized as an ellipse.

Note. The quadratic kind can be determined computing the characteristic polynomial roots. In this case

$M = \left(\begin{matrix}13 & 3 \sqrt{3} \\ 3 \sqrt{3} & 7\end{matrix}\right)$

has the characteristic polynomial

${\lambda}^{2} - \text{trace} \left(M\right) \lambda + \det \left(M\right) = 0$

or

${\lambda}^{2} - 20 \lambda + 64 = 0$ with roots

$\lambda = \left\{4 , 16\right\}$ characteririzing an ellipse.