Answer
Graph:
.
Work Step by Step
$ r=\displaystyle \frac{ke}{1\pm e\sin\theta}\qquad$
The main axis is vertical.
We want $1\pm...$ in the denominator,
$\displaystyle \frac{4\div 2}{(2-\sin\theta)\div 2}$=$\displaystyle \frac{2}{1-\frac{1}{2}\sin\theta}$
$ r=\displaystyle \frac{4(\frac{1}{2})}{1-\frac{1}{2}\sin\theta}\qquad$
Compare with $r=\displaystyle \frac{ke}{1\pm e\sin\theta}$
$e=\displaystyle \frac{1}{2}, \qquad$
This is an ellipse with vertical major axis.
$ k=4,\quad y=-4$ is a directrix
$a(1-e^{2})=ke$
$a\displaystyle \cdot\frac{3}{4}=2$
$a=\displaystyle \frac{8}{3}$
$c=ae=\displaystyle \frac{4}{3}$
Center is at $\displaystyle \frac{4}{3}$ units above the focus at the origin; polar coordinates are $(\displaystyle \frac{4}{3},\frac{\pi}{2}).$
The vertices are at $\displaystyle \frac{8}{3}$ units above and below the center,
$(\displaystyle \frac{8}{3}+\frac{4}{3},\frac{\pi}{2})=(4,\frac{\pi}{2})$
$(\displaystyle \frac{4}{3}-\frac{8}{3},\frac{\pi}{2})=(-\frac{4}{3},\frac{\pi}{2})=(\frac{4}{3},\frac{3\pi}{2})$
