How do you use rotation of axes to identify and sketch the curve of sqrt3xy+y^2=1?

Aug 27, 2016

$\frac{3}{2} {X}^{2} - {Y}^{2} = 1$

Explanation:

This quadratic can be written as

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

where

$p = \left(x , y\right)$ and $M = \left(\begin{matrix}0 & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & 1\end{matrix}\right)$

Choosing a rotation given by

$R \left(\theta\right) = \left(\begin{matrix}C o s \theta & - S \int h \eta \\ S \int h \eta & C o s \theta\end{matrix}\right)$

and a new set of coordinates

$P = \left(X , Y\right) = p . R \left(\theta\right)$ then

$P . {R}^{- 1} \left(\theta\right) . M . R \left(\theta\right) . {P}^{T} = 1$

with

R^(-1)(theta).M.R(theta) = ((Sintheta (-sqrt[3] Costheta + Sintheta), 1/2 (sqrt[3] Cos(2theta) - Sin(2 theta))),(1/ 2 (sqrt[3] Cos(2theta) - Sin(2 theta)), Cos theta (Costheta + sqrt[3] Sintheta)))

Choosing $\theta$ such that

$\frac{1}{2} \left(\sqrt{3} C o s \left(2 \theta\right) - S \in \left(2 \theta\right)\right) = 0$ for

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

we have

$\frac{3}{2} {X}^{2} - {Y}^{2} = 1$

which is a hyperbola.