# How do you convert x^2/4 + y^2 = 4 into polar form?

Oct 20, 2016

$r = \frac{4}{\sqrt{{\cos}^{2} \theta + 4 {\sin}^{2} \theta}}$

#### Explanation:

Let $x = r \cos \theta$ and $y = r \sin \theta$
then ${x}^{2} / 4 + {y}^{2} = 4$
$\implies$ $\frac{{r}^{2} {\cos}^{2} \theta}{4} + {r}^{2} {\sin}^{2} \theta = 4$
Simplifying
${r}^{2} {\cos}^{2} \theta + 4 {r}^{2} {\sin}^{2} \theta = 16$
${r}^{2} \left({\cos}^{2} \theta + 4 {\sin}^{2} \theta\right) = 16$
${r}^{2} = \frac{16}{{\cos}^{2} \theta + 4 {\sin}^{2} \theta}$
$r = \frac{4}{\sqrt{{\cos}^{2} \theta + 4 {\sin}^{2} \theta}}$