# How do you convert R=6/(2+cos(theta)) into cartesian form?

Nov 28, 2017

$3 {x}^{2} + 12 x + 4 {y}^{2} - 36 = 0$

#### Explanation:

$r = \frac{6}{2 + \cos \left(\theta\right)}$

$2 r + r \cos \left(\theta\right) = 6$

For polar coordinates; $r$ is equal to $\sqrt{{x}^{2} + {y}^{2}}$ and $r \cos \left(\theta\right)$ is equal to $x$ in cartesian ones. Hence,

$2 \sqrt{{x}^{2} + {y}^{2}} + x = 6$

$2 \sqrt{{x}^{2} + {y}^{2}} = 6 - x$

$4 \cdot \left({x}^{2} + {y}^{2}\right) = {\left(6 - x\right)}^{2}$

$4 {x}^{2} + 4 {y}^{2} = {x}^{2} - 12 x + 36$

$3 {x}^{2} + 12 x + 4 {y}^{2} - 36 = 0$