# How do you convert r(2 + cos theta) = 1 into cartesian form?

Nov 26, 2016

The equation is $3 {x}^{2} + 4 {y}^{2} + 2 x - 1 = 0$

#### Explanation:

To transform from polar coordinates $\left(r , \theta\right)$ into rectangular coordinates $\left(x , y\right)$, we use the folowing

$x = r \cos \theta$

$y = r \sin \theta$

and ${x}^{2} + {y}^{2} = {r}^{2}$

Therefore,

$r \left(2 + \cos \theta\right) = 1$

$2 r + r \cos \theta = 1$

$2 \sqrt{{x}^{2} + {y}^{2}} + x = 1$

$2 \sqrt{{x}^{2} + {y}^{2}} = 1 - x$

Squaring both sides,

$4 \left({x}^{2} + {y}^{2}\right) = {\left(1 - x\right)}^{2} = 1 - 2 x + {x}^{2}$

$3 {x}^{2} + 4 {y}^{2} + 2 x - 1 = 0$

Nov 26, 2016

It is the ellipsys $3 {x}^{2} + 4 {y}^{2} + 2 x - 1 = 0$

#### Explanation:

$\rho = \sqrt[2]{{x}^{2} + {y}^{2}}$ and $x = \rho \cdot \sin \theta$ and $y = \rho \cdot \cos \theta$
$\sqrt[2]{{x}^{2} + {y}^{2}} \cdot \left(2 + \frac{x}{\sqrt[2]{{x}^{2} + {y}^{2}}}\right) = 1$
that becomes
$2 \sqrt[2]{{x}^{2} + {y}^{2}} + x = 1$
or
${x}^{2} + {y}^{2} = {\left(\frac{1 - x}{2}\right)}^{2}$
or
${x}^{2} + {y}^{2} - {\left(1 - x\right)}^{2} / 4 = 0$