# How do you graph the conic x^2-4xy+y^2+1=0 by first rotations the axis and eliminating the xy term?

Jan 28, 2017

#### 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 ${x}^{2} - 4 x y + {y}^{2} + 1 = 0$, we have $A = C = 1$ and $B = - 4$ and hence $\cot 2 \theta = 0$ i.e. $\theta = \frac{\pi}{4}$

Hence relation is give by $x = x ' \cos \left(\frac{\pi}{4}\right) - y ' \sin \left(\frac{\pi}{4}\right)$ and $y = x ' \sin \left(\frac{\pi}{4}\right) + y ' \cos \left(\frac{\pi}{4}\right)$ i.e.

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

Hence, we get ${\left(\frac{x '}{\sqrt{2}} - \frac{y '}{\sqrt{2}}\right)}^{2} - 4 \left(\frac{x '}{\sqrt{2}} - \frac{y '}{\sqrt{2}}\right) \left(\frac{x '}{\sqrt{2}} + \frac{y '}{\sqrt{2}}\right) + {\left(\frac{x '}{\sqrt{2}} + \frac{y '}{\sqrt{2}}\right)}^{2} + 1 = 0$

or $\left(\frac{x {'}^{2}}{2} + \frac{y {'}^{2}}{2} - x ' y '\right) - 4 \left(\frac{x {'}^{2}}{2} - \frac{y {'}^{2}}{2}\right) + \left(\frac{x {'}^{2}}{2} + \frac{y {'}^{2}}{2} + x ' y '\right) + 1 = 0$

or $- x {'}^{2} + 3 y {'}^{2} + 1 = 0$ or $x {'}^{2} - 3 y {'}^{2} = 1$

The two graphs are as follows:
graph{x^2-4xy+y^2+1=0 [-10, 10, -5, 5]}
and
graph{x^2-3y^2=1 [-10, 10, -5, 5]}