# Question #a63a9

Jan 14, 2018

$r = \frac{1}{5} \sec \left(\theta\right) \tan \left(\theta\right)$

#### Explanation:

To convert to polar let:

$x = r \cos \left(\theta\right)$
$y = r \sin \left(\theta\right)$

and then preferably solve for $r$.

For $y = 5 {x}^{2}$ we have:

$r \sin \left(\theta\right) = 5 {\left(r \cos \left(\theta\right)\right)}^{2}$

$r \sin \left(\theta\right) = 5 {r}^{2} {\cos}^{2} \left(\theta\right)$

Since we know that $r$ is not universally 0, we can divide through by $r$:

$\sin \left(\theta\right) = 5 r {\cos}^{2} \left(\theta\right)$

now we divide both sides by $5 {\cos}^{2} \left(\theta\right)$:

$\sin \frac{\theta}{5 {\cos}^{2} \left(\theta\right)} = r$

Since $\sin \frac{\theta}{\cos} \left(\theta\right) = \tan \left(\theta\right)$ and $\frac{1}{\cos} \left(\theta\right) = \sec \left(\theta\right)$, we can rewrite one more time:

$r = \frac{1}{5} \sec \left(\theta\right) \tan \left(\theta\right)$