Answer
(a) See graph and explanations.
(b) directrix $x=\frac{6}{7}$. vertices $V_1(\frac{2}{3},0)$ and $V_2(-\frac{6}{5}, \pi)$
(c) center $C(\frac{14}{15}, 0)$.
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 $r=\frac{3}{1+\frac{7}{2}cos\theta}$, we can identify that $e=\frac{7}{2}\gt1$ by comparing it with a standard equation in step-1. Thus the graph of the equation will be a hyperbola with a directrix to the right of the origin.
Step 3. See graph.
(b) With $e=7/2$ and $ed=3$, we have $d=\frac{6}{7}$, we can identify that the directrix as $x=d=\frac{6}{7}$.
To find the vertices, let $\theta=0, \pi$, we have $r(0)=\frac{2}{3}$ and $r(\pi)=-\frac{6}{5}$, thus we have the vertices $V_1(\frac{2}{3},0)$ and $V_2(-\frac{6}{5}, \pi)$ as indicated in the figure.
(c) The center is half way between the two vertices, thus we have $r(c)=\frac{2}{3}+(\frac{6}{5}-\frac{2}{3})/2=\frac{14}{15}$ and the center is at $C(\frac{14}{15}, 0)$.
The asymptotes are shown in the figure
