# What is 6y+y^2=x^2 in polar form?

## $6 y + {y}^{2} = {x}^{2}$ in polar form. I know the answer is $r = \frac{6 \sin \theta}{\cos 2 \theta}$ but I do not know how to get there.

Apr 24, 2018

Put $x = r \cos \theta$ and $y = r \sin \theta$

and use the property ${\cos}^{2} \theta - {\sin}^{2} \theta = \cos 2 \theta$

#### Explanation:

$6 y + {y}^{2} = {x}^{2}$........................ (given equation)

Put $x = r \cos \theta$ and $y = r \sin \theta$ , we get :-

$6 r \sin \theta + {r}^{2} {\sin}^{2} \theta = {r}^{2} {\cos}^{2} \theta$

$\Rightarrow 6 r \sin \theta = {r}^{2} {\cos}^{2} \theta - {r}^{2} {\sin}^{2} \theta$

$\Rightarrow 6 r \sin \theta = {r}^{2} \left({\cos}^{2} \theta - {\sin}^{2} \theta\right)$

$\Rightarrow 6 r \sin \theta = {r}^{2} \cos 2 \theta$..............$\left\{{\cos}^{2} \theta - {\sin}^{2} \theta = \cos 2 \theta\right\}$

$\Rightarrow 6 \sin \theta = r \cos 2 \theta$

$\therefore r = \frac{6 \sin \theta}{\cos 2 \theta}$ is the Polar form of $6 y + {y}^{2} = {x}^{2}$