Answer
See image:
Work Step by Step
We have a sine in the denominator, so the main axis is vertical.
We want $1\pm...$ in the denominator,
$\displaystyle \frac{400\div 16}{(16+8\sin\theta)\div 16}$=$\displaystyle \frac{25}{1+(\frac{1}{2})\sin\theta}$
$r=\displaystyle \frac{50(\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 a vertical major axis.
$k=50,\quad y=50$ is a directrix
$a(1-e^{2})=ke$
$a\displaystyle \cdot\frac{3}{4}=25$
$a=\displaystyle \frac{100}{3}$
$c=ae=\displaystyle \frac{50}{3}$
The center is at $\displaystyle \frac{50}{3}$ units below the focus at the origin, with polar coordinates $(\displaystyle \frac{50}{3},\frac{3\pi}{2}).$
The vertices are at $\displaystyle \frac{100}{3}$ units above and below the center,
$(\displaystyle \frac{50}{3}+\frac{100}{3},\frac{3\pi}{2})=(50,\frac{3\pi}{2})$
$(\displaystyle \frac{50}{3}-\frac{100}{3},\frac{3\pi}{2})=(-\frac{50}{3},\frac{3\pi}{2})=(\frac{50}{3},\frac{\pi}{2})$
