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# How do you rotate the axes to transform the equation x^2+2sqrt3xy-y^2=7 into a new equation with no xy term and then find the angle of rotation?

Apr 23, 2017

$2 {x}^{2} - 2 {y}^{2} - 7 = 0$

#### Explanation:

A conic equation of the type of $A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$ is rotated by an angle $\theta$, to form a new Cartesian plane with coordinates $\left(x ' , y '\right)$, if $\theta$ is appropriately chosen, we can have a new equation without term $x y$ i.e. of standard form.

The relation between coordinates $\left(x , y\right)$ and $\left(x ' . y '\right)$ can be expressed as
$x = x ' \cos \theta - y ' \sin \theta$ and $y = x ' \sin \theta + y ' \cos \theta$

or $x ' = x \cos \theta + y \sin \theta$ and $y = - x \sin \theta + y \cos \theta$

for this we need to have $\theta$ given by $\cot 2 \theta = \frac{A - C}{B}$

In the given case as equation is ${x}^{2} + 2 \sqrt{3} x y - {y}^{2} = 7$, we have $A = 1$, $B = 2 \sqrt{3}$ and $C = - 1$ and hence $\cot 2 \theta = \frac{2}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}$ i.e. $\theta = \frac{\pi}{6}$

Hence relation is give by $x = x ' \cos \left(\frac{\pi}{6}\right) - y ' \sin \left(\frac{\pi}{6}\right)$ and $y = x ' \sin \left(\frac{\pi}{6}\right) + y ' \cos \left(\frac{\pi}{6}\right)$ i.e.

$x = \frac{x ' \sqrt{3}}{2} - \frac{y '}{2}$ and $y = \frac{x '}{2} + \frac{y ' \sqrt{3}}{2}$

and ${x}^{2} + 2 \sqrt{3} x y - {y}^{2} = 7$ is

${\left(\frac{x ' \sqrt{3}}{2} - \frac{y '}{2}\right)}^{2} + 2 \sqrt{3} \left(\frac{x ' \sqrt{3}}{2} - \frac{y '}{2}\right) \left(\frac{x '}{2} + \frac{y ' \sqrt{3}}{2}\right) - {\left(\frac{x '}{2} + \frac{y ' \sqrt{3}}{2}\right)}^{2} = 7$

which simplifies to $2 \left(x {'}^{2} - y {'}^{2}\right) = 7$ or we can say

$2 {x}^{2} - 2 {y}^{2} - 7 = 0$

graph{(2x^2-2y^2-7)(x^2+2sqrt3xy-y^2-7)=0 [-10, 10, -5, 5]}