# Question #104ce

Apr 4, 2017

${y}^{2} = 3 \sqrt{{x}^{2} + {y}^{2}} + 3 x - {x}^{2}$

#### Explanation:

Use the following formulas (given by Pythagorean theorem, definition of sine, and definition of cosine, respectively):
$\textcolor{b l u e}{{x}^{2} + {y}^{2} = {r}^{2}}$
$\textcolor{b l u e}{y = r \sin \theta}$
$\textcolor{b l u e}{x = r \cos \theta}$

We have to manipulate the equation so that it is possible to substitute the formulas in.

(Note that the graph of $r = 3 + 3 \cos \theta$ is a cardioid, which is not a conic section, so the rectangular form of the equation will not be in any standard form.)

$r = 3 + 3 \cos \theta$

$\textcolor{b l u e}{{r}^{2}} = 3 r + 3 \textcolor{b l u e}{r \cos \theta}$

${x}^{2} + {y}^{2} = 3 \sqrt{{x}^{2} + {y}^{2}} + 3 x$

${y}^{2} = 3 \sqrt{{x}^{2} + {y}^{2}} + 3 x - {x}^{2}$