# How do you convert r=4tanθ into cartesian form?

Jul 19, 2016

${x}^{4} + {x}^{2} {y}^{2} - 16 {y}^{2} = 0$

#### Explanation:

The relation between polar coordinates $\left(r , \theta\right)$ and Cartesian coordinates $\left(x , y\right)$ is given by $x = r \cos \theta$, $y = r \sin \theta$, $\tan \theta = \frac{y}{x}$ and ${r}^{2} = {x}^{2} + {y}^{2}$.

Hence, $r = 4 \tan \theta$

$\Leftrightarrow {r}^{2} = 16 {\tan}^{2} \theta$

$\Leftrightarrow {x}^{2} + {y}^{2} = 16 \left({y}^{2} / {x}^{2}\right)$

or ${x}^{4} + {x}^{2} {y}^{2} - 16 {y}^{2} = 0$