# How do you graph r(2 + cos theta) = 1?

Mar 31, 2016

This is polar equation$\frac{\frac{1}{2}}{r} = 1 + \left(\frac{1}{2}\right) \cos \theta$ of an ellipse referred to a focus as pole and major axis (in the direction of nearer end) as initial line. Major axis = 4/3 and eccentricity = 1/2..

#### Explanation:

Compare with the standard form $\frac{l}{r} = 1 + e \cos \theta$, the rearranged equation .$\frac{\frac{1}{2}}{r} = 1 + \left(\frac{1}{2}\right) \cos \theta$

$e = \frac{1}{2} \mathmr{and} l = a \left(1 - {e}^{2}\right) = \frac{1}{2}$.
$a = \frac{1}{2 \left(1 - \frac{1}{4}\right)} = \frac{2}{3}$.
O(0, 0) is a focus S.
The other focus S'$\left(2 a e , \pi\right) = \left(\frac{2}{3} , \pi\right)$
The center C is the midpoint $\left(\frac{1}{3} , \pi\right)$, in-between.
Put $\theta = 0 , \pi$ in the equation to get ends of the major axis,
$A \left(\frac{1}{3} , 0\right) , A ' \left(1 , \pi\right)$

Put $r = a = \frac{2}{3}$ to get the ends of the minor axis, B(2/3, 2pi/3), B'(2/3, -2pi/3)#.