Answer
(a) See graph and explanations.
(b) directrix $x=4$, vertices $V_1(\frac{8}{3},0)$ and $V_2(-8, \pi)$
(c) center $C(\frac{16}{3}, 0)$.
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. Write the equation given in the Exercise as $r=\frac{8}{1+2cos\theta}$, we can identify that $e=2\gt1$ 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 $e=2$ and $ed=8$, we have $d=4$, we can identify that the directrix as $x=d=4$. To find the vertices, let $\theta=0, \pi$, we have $r(0)=\frac{8}{3}$ and $r(\pi)=-8$, thus we have the vertices $V_1(\frac{8}{3},0)$ and $V_2(-8, \pi)$ as indicated in the figure.
(c) The center is half way between the two vertices, thus we have $r(c)=\frac{8}{3}+(8-\frac{8}{3})/2=\frac{16}{3}$ and the center is at $C(\frac{16}{3}, 0)$.
The asymptotes are shown in the figure.
