# How do you use rotation of axes to identify and sketch the curve of 97x^2+192xy+153y^2=225?

Feb 15, 2018

The curve is an ellipse and after rotation it is $9 {x}^{2} + {y}^{2} = 9$

#### 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 $97 {x}^{2} + 192 x y + 153 {y}^{2} - 225 = 0$, we have $A - C = - 56$ and $B = 192$ and hence $\cot 2 \theta = - \frac{7}{24}$ i.e. $\frac{{\cot}^{2} \theta - 1}{2 \cot \theta} = - \frac{7}{24}$ or $12 {\cot}^{2} \theta + 7 \cot \theta - 12 = 0$ or $\left(4 \cot \theta - 3\right) \left(3 \cot \theta + 4\right) = 0$ i.e. $\cot \theta = \frac{3}{4}$ or $\cot \theta = - \frac{4}{3}$.

We consider only acute angle i.e. $\cot \theta = \frac{3}{4}$, which leads to $\cos \theta = \frac{3}{5}$ and $\sin \theta = \frac{4}{5}$

Hence relation is give by $x = \frac{3}{5} x ' - \frac{4}{5} y '$ and $y = \frac{4}{5} x ' + \frac{3}{5} y '$

Hence, we get $97 {\left(\frac{3 x '}{5} - \frac{4 y '}{5}\right)}^{2} + 192 \left(\frac{3 x '}{5} - \frac{4 y '}{5}\right) \left(\frac{4 x '}{5} + \frac{3 y '}{5}\right) + 153 {\left(\frac{4 x '}{5} + \frac{3 y '}{5}\right)}^{2} - 225 = 0$

or $97 \left(\frac{9 x {'}^{2}}{25} + \frac{16 y {'}^{2}}{25} - \frac{24 x ' y '}{25}\right) + 192 \left(\frac{12 x {'}^{2}}{25} - \frac{12 y {'}^{2}}{25} - \frac{7 x ' y '}{25}\right) + 153 \left(\frac{16 x {'}^{2}}{25} + \frac{9 y {'}^{2}}{25} + \frac{24 x ' y '}{25}\right) - 225 = 0$

or $225 x {'}^{2} + 25 y {'}^{2} - 225 = 0$ or $9 x {'}^{2} + 1 y {'}^{2} = 9$

The two graphs are as follows:
graph{97x^2+192xy+153y^2-225=0 [-10, 10, -5, 5]}
and
graph{9x^2+y^2=9 [-10, 10, -5, 5]}