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

Feb 8, 2017

${\left(x + \frac{1}{3}\right)}^{2} / \left(\frac{3}{4}\right) + {y}^{2} / \left(\frac{1}{3}\right) = 1$ See the markings at the center and the pole of the graph of this ellipse.

#### Explanation:

graph{(2sqrt(x^2+y^2)+x-1)(x^2+y^2-.0005)((x+1/3)^2+y^2-.0005)=0 [-1.25, 1.25, -0.625, 0.625]} The equation has the form

$\frac{\frac{1}{2}}{r} = 1 + \frac{1}{2} \cos \theta$.

This represents the ellipse with focus S as pole, $\theta = 0$ for S-side

major axis, eccentricity $e = \frac{1}{2}$ and semi major axis a = 2/3, from

$a \left(1 - {e}^{2}\right) = l = \frac{1}{2}$.

The conversion formula is

$r \left(\cos \theta , \sin \theta\right) = \left(x , y\right)$.

So, $2 r + r \cos \theta = 2 \sqrt{{x}^{2} + {y}^{2}} + x = 1$, giving

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

In the standard form, this is

${\left(x + \frac{1}{3}\right)}^{2} / \left(\frac{3}{4}\right) + {y}^{2} / \left(\frac{1}{3}\right) = 1$

Feb 8, 2017

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

#### Explanation:

In converting from polar $\left(r , \theta\right)$ to Cartesian $\left(x , y\right)$
we have the relations:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{r = \sqrt{{x}^{2} + {y}^{2}}}$
and
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{\cos \left(\theta\right) = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}}$
(and others)

Polar form: $\textcolor{red}{r} \left(2 + \textcolor{b l u e}{\cos \left(\theta\right)}\right) = 1$
becomes (in Cartesian form)
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{\sqrt{{x}^{2} + {y}^{2}}} \cdot \left(2 + \textcolor{b l u e}{\frac{x}{\sqrt{{x}^{2} + {y}^{2}}}}\right) = 1$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 2 \sqrt{{x}^{2} + {y}^{2}} + x = 1$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow \sqrt{{x}^{2} + {y}^{2}} = \frac{1 - x}{2}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {x}^{2} + {y}^{2} = \frac{1 - 2 x + {x}^{2}}{4}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 4 {x}^{2} + 4 {y}^{2} = 1 - 2 x + {x}^{2}$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow 3 {x}^{2} + 4 {y}^{2} + 2 x = 1$