How do you convert 2x+y^2 = 5 to polar form?

Nov 13, 2016

Explanation:

Substitute $\left(r \cos \left(\theta\right)\right)$ for x and ${r}^{2} {\sin}^{2} \left(\theta\right)$ for ${y}^{2}$:

${r}^{2} {\sin}^{2} \left(\theta\right) +$2rcos(theta) - 5 = 0#

Use the positive root of the quadratic formula:

$r = \frac{- 2 \cos \left(\theta\right) + \sqrt{4 {\cos}^{2} \left(\theta\right) + 20 {\sin}^{2} \left(\theta\right)}}{2 {\sin}^{2} \left(\theta\right)}$

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

$r = \frac{- \cos \left(\theta\right) + \sqrt{1 + 4 {\sin}^{2} \left(\theta\right)}}{{\sin}^{2} \left(\theta\right)}$