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# How do you rotate the axes to transform the equation x^2+xy=3 into a new equation with no xy term and then find the angle of rotation?

Feb 1, 2017

$\frac{x {'}^{2}}{4} \left(2 + 2 \sqrt{2}\right) + \frac{y {'}^{2}}{4} \left(2 - 2 \sqrt{2}\right) - 3 = 0$

#### 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 ${x}^{2} + x y - 3 = 0$, we have $A = 1$ and $B = 1$ and $C = 0$, hence $\cot 2 \theta = 1$ i.e. $\theta = \frac{\pi}{8}$

Hence relation is give by $x = x ' \cos \left(\frac{\pi}{8}\right) - y ' \sin \left(\frac{\pi}{8}\right)$ and $y = x ' \sin \left(\frac{\pi}{8}\right) + y ' \cos \left(\frac{\pi}{8}\right)$ i.e.

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

Hence, we get ${\left(\frac{x ' \sqrt{2 + \sqrt{2}}}{2} - \frac{y ' \sqrt{2 - \sqrt{2}}}{2}\right)}^{2} + \left(\frac{x ' \sqrt{2 + \sqrt{2}}}{2} - \frac{y ' \sqrt{2 - \sqrt{2}}}{2}\right) \left(\frac{x ' \sqrt{2 - \sqrt{2}}}{2} + \frac{y ' \sqrt{2 + \sqrt{2}}}{2}\right) - 3 = 0$

or $\left(\frac{x {'}^{2} \left(2 + \sqrt{2}\right)}{4} + \frac{y {'}^{2} \left(2 - \sqrt{2}\right)}{4} - \frac{2 x ' y ' \sqrt{2}}{4}\right) + \left(\frac{x {'}^{2} \sqrt{2}}{4} - \frac{y {'}^{2} \sqrt{2}}{4} + \frac{x ' y ' \left(2 + \sqrt{2}\right)}{4} - \frac{x ' y ' \left(2 - \sqrt{2}\right)}{4}\right) - 3 = 0$

or $\frac{x {'}^{2}}{4} \left(2 + 2 \sqrt{2}\right) + \frac{y {'}^{2}}{4} \left(2 - 2 \sqrt{2}\right) - 3 = 0$

The two graphs are as follows:
graph{x^2+xy-3=0 [-10, 10, -5, 5]}
and
graph{(x^2)/4(2+2sqrt2)+(y^2)/4(2-2sqrt2)-3=0 [-10, 10, -5, 5]}