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,
$\displaystyle \frac{8\div 2}{(2-2\sin\theta)\div 2}=\frac{4}{1-\sin\theta}=\frac{-4(-1)}{1+(-1)\sin\theta}$
Compare with $r=\displaystyle \frac{ke}{1\pm e\sin\theta}$
$e=1$
Hence, this is a parabola.
$k=4$
$y=-4$ is the directrix, so the parabola opens up
The vertex is halfway between the directrix and focus, 2 units below the focus at the origin, on the y-axis $(\theta=3\pi/2)$ with polar coordinates: $(2,3\pi/2).$
