Answer
(a) See graph and explanations.
(b) directrix $y=-\frac{20}{3}$. vertices $V_1(-20, \frac{\pi}{2})$ and $V_2(4, \frac{3\pi}{2})$
(c) center $C(12, \frac{3\pi}{2})$.
Work Step by Step
(a) Step 1. The standard form of a conic polar equation is $r=\frac{ed}{1\pm e\cdot cos\theta}$ or $r=\frac{ed}{1\pm e\cdot sin\theta}$. The graph of the equation will be a hyperbola if $e\gt1$
Step 2. Rewrite the equation given in the Exercise as $r=\frac{10}{1-\frac{3}{2}sin\theta}$, we can identify that $e=\frac{3}{2}\gt1$ by comparing it with a standard equation in step-1. Thus the graph of the equation will be a hyperbola with a directrix below the origin.
Step 3. See graph.
(b) With $e=\frac{3}{2}$ and $ed=10$, we have $d=\frac{20}{3}$, we can identify that the directrix as $y=-d=-\frac{20}{3}$.
To find the vertices, let $\theta=\frac{\pi}{2},\frac{3\pi}{2}$, we have $r(\frac{\pi}{2})=-20$ and $r(\frac{3\pi}{2})=4$, thus we have the vertices $V_1(-20, \frac{\pi}{2})$ and $V_2(4, \frac{3\pi}{2})$ as indicated in the figure.
(c) The center is half way between the two vertices, thus we have $r(c)=4+(20-4)/2=12$ and the center is at $C(12, \frac{3\pi}{2})$.
The asymptotes are shown in the figure
