Answer
(a) See graph and explanations.
(b) directrix $y=-\frac{5}{2}$. vertices $V_1(-\frac{10}{3},\frac{\pi}{2})$ and $V_2(2, \frac{3\pi}{2})$
(c) center $C(\frac{8}{3}, \frac{3\pi}{2})$.
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. Write the equation given in the Exercise as $r=\frac{10}{1-4sin\theta}$, we can identify that $e=4\gt1$ 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 $e=4$ and $ed=10$, we have $d=\frac{5}{2}$, we can identify that the directrix as $y=-d=-\frac{5}{2}$.
To find the vertices, let $\theta=\pi/2, 3\pi/2$, we have $r(\pi/2)=-\frac{10}{3}$ and $r(3\pi/2)=2$, thus we have the vertices $V_1(-\frac{10}{3},\frac{\pi}{2})$ and $V_2(2, \frac{3\pi}{2})$ as indicated in the figure.
(c) The center is half way between the two vertices, thus we have $r(c)=2+(\frac{10}{3}-2)/2=\frac{8}{3}$ and the center is at $C(\frac{8}{3}, \frac{3\pi}{2})$.
The asymptotes are shown in the figure
