# How do you graph r(2 - costheta) = 2?

May 3, 2016

In Catrersian form it is $3 {x}^{2} + 4 {y}^{2} - 4 x - 4 = 0$ an ellipse.

#### Explanation:

If $\left(r , \theta\right)$ is in polar form and $\left(x , y\right)$ in Cartesian form the relation between them is as follows:

$x = r \cos \theta$, $y = r \sin \theta$, ${r}^{2} = {x}^{2} + {y}^{2}$ and $\tan \theta = \frac{y}{x}$

Or, $\cos \theta = \frac{x}{r}$, $\sin \theta = \frac{y}{r}$, $\theta = {\tan}^{- 1} \left(\frac{y}{x}\right)$ and $\cot \theta = \frac{x}{y}$.

Hence, $r \left(2 - \cos \theta\right) = 2$ can be written as

$2 r - r \cos \theta = 2$

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

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

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

$4 {x}^{2} + 4 {y}^{2} = 4 + {x}^{2} + 4 x$ or

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

As coefficients of ${x}^{2}$ and ${y}^{2}$ are both positive but not equal, this is an ellipse.

The above can be written as

$3 \left({x}^{2} - \frac{4}{3} x + \frac{4}{9}\right) + 4 {y}^{2} - 4 - \frac{12}{9} - 0$

or $3 {\left(x - \frac{2}{3}\right)}^{2} + 4 {\left(y - 0\right)}^{2} = \frac{48}{9} = \frac{16}{3}$

or $\frac{9}{16} {\left(x - \frac{2}{3}\right)}^{2} + \frac{3}{4} {\left(y - 0\right)}^{2} = 1$

or ${\left(x - \frac{2}{3}\right)}^{2} / \left(\frac{16}{9}\right) + {\left(y - 0\right)}^{2} / \left(\frac{4}{3}\right) = 1$

Center of ellipse is $\left(\frac{2}{3} , 0\right)$

Major axis is $2 \times \frac{4}{3} = \frac{8}{3}$ and minor axis is $2 \times \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}}$

graph{3x^2+4y^2-4x-4=0 [-3, 3, -1.5, 1.5]}