How do you graph the conic 2x^2-3xy-2y^2+10=0 by first rotations the axis and eliminating the xy term?

Sep 15, 2016

${X}^{2} - {Y}^{2} + 4 = 0$ which is a hyperbola.

Explanation:

With the rotation

${R}_{\theta} = \left(\begin{matrix}c \theta & - s \theta \\ s \theta & c \theta\end{matrix}\right)$, calling $p = \left(\begin{matrix}x \\ y\end{matrix}\right)$ and $P = \left(\begin{matrix}X \\ Y\end{matrix}\right)$ we have first

${p}^{T} M p + 10 = 0$ representing the conic, with $M = \left(\begin{matrix}2 & - \frac{3}{2} \\ - \frac{3}{2} & - 2\end{matrix}\right)$. Here ${.}^{T}$ represents transposition. Now introducing a change of coordinates

${R}_{\theta} P = p$ we will have

${P}^{T} {R}_{\theta}^{T} M {R}_{\theta} P + 10 = 0$

Calling M_(theta) = R_(theta)^TMR_(theta)=((2 Cos(2 theta) - 3 Cos(theta) Sin(theta), -3/2Cos(2 theta) - 2 Sin(2 theta)),(-3/2 Cos(2 theta) - 2 Sin(2 theta), -2 Cos(2 theta) + 3 Cos(theta) Sin(theta)))

Now, choosing $\theta$ such that $- \frac{3}{2} C o s \left(2 \theta\right) - 2 S \in \left(2 \theta\right) = 0$ we obtain

$\tan \left(2 \theta\right) = - \frac{3}{4}$ and

$\theta = \frac{1}{2} \left(- 0.643501 \pm k \pi\right)$ Choosing for $k = 0$ we have

$\theta = - 0.321751$ and

${M}_{\theta} = \left(\begin{matrix}\frac{5}{2} & 0 \\ 0 & - \frac{5}{2}\end{matrix}\right)$ and in the new coordinates, the conic reads

$\frac{5}{2} {X}^{2} - \frac{5}{2} {Y}^{2} + 10 = 0$ or

${X}^{2} - {Y}^{2} + 4 = 0$ which is a hyperbola.