# How do you convert y=x-2y+xy^2  into a polar equation?

May 21, 2018

$r = \frac{\cos \theta - \sin \theta}{\left(2 - \cos \theta\right) \sin \theta \cos \theta}$

#### Explanation:

$y = x - 2 y + x {y}^{2}$
$x = r \cos \theta$
$y = r \sin \theta$
Thus,
$r \sin \theta = r \cos \theta - 2 r \cos \theta \times r \sin \theta + r \cos \theta \times {\left(r \sin \theta\right)}^{2}$

$r \sin \theta = r \cos \theta - 2 {r}^{2} \sin \theta \cos \theta + {r}^{2} {\sin}^{2} \theta \cos \theta$

$r \sin \theta - r \cos \theta = {r}^{2} \left(- 2 \sin \theta \cos \theta + {\sin}^{2} \theta \cos \theta\right)$

$r \left(\sin \theta - \cos \theta\right) = {r}^{2} \sin \theta \cos \theta \left(- 2 + \cos \theta\right)$

$r \left(\sin \theta - \cos \theta\right) + {r}^{2} \sin \theta \cos \theta \left(2 - \cos \theta\right)$

$r \left(\sin \theta - \cos \theta + r \sin \theta \cos \theta \left(2 - \cos \theta\right)\right) = 0$

$r = 0$

$\sin \theta - \cos \theta + r \sin \theta \cos \theta \left(2 - \cos \theta\right) = 0$

$r \sin \theta \cos \theta \left(2 - \cos \theta\right) = \cos \theta - \sin \theta$
$r = \frac{\cos \theta - \sin \theta}{\left(2 - \cos \theta\right) \sin \theta \cos \theta}$