Answer
(a) See graph and explanations.
(b) directrix $x=\frac{5}{3}$. vertex $(\frac{5}{6},0)$
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 parabola if $e=1$
Step 2. Rewrite the equation given in the Exercise as $r=\frac{5/3}{1+cos\theta}$, we can identify that $e=1$ by comparing it with a standard equation in step-1. Thus the graph of the equation will be a parabola (open to the left).
Step 3. See graph.
(b) With $e=1$ and $ed=\frac{5}{3}$, we have $d=\frac{5}{3}$, we can identify that the directrix is $x=d=\frac{5}{3}$. The focus is at the origin and the vertex is at $(\frac{5}{6},0)$ which is half way between the focus and the directrix as indicated in the graph.
