Answer
(a) See graph and explanations
(b) directrix $x=-4$. vertices $V_1(4,0)$ and $V_2(\frac{4}{3}, \pi)$
(c) center $C(\frac{4}{3}, 0)$. length of the major axes $\frac{16}{3}$. length of the minor axes $\frac{8\sqrt 3}{3}$
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 $r=\frac{2}{1-\frac{1}{2}cos\theta}$, we can identify that $e=\frac{1}{2}$ by comparing it with a standard equation in step-1. Thus the graph of the equation will be an ellipse with a directrix to the left of the curve.
Step 3. See graph.
(b) With $e=\frac{1}{2}$ and $ed=2$, we have $d=4$, we can identify that the directrix as $x=-d=-4$. To find the vertices, let $\theta=0, \pi$, we have $r(0)=4$ and $r(\pi)=\frac{4}{3}$, thus we have the vertices $V_1(4,0)$ and $V_2(\frac{4}{3}, \pi)$ as indicated in the figure.
(c) The center of the ellipse is half way between the two vertices, thus we have $r(c)=\frac{4-4/3}{2}=\frac{4}{3}$ and the center is at $C(\frac{4}{3}, 0)$.
The length of the major axes is the distance between the two vertices, thus $2a=4+\frac{4}{3}=\frac{16}{3}$.
With $a=\frac{8}{3}, c=\frac{4}{3}$ (origin is one of the focus), we have $b=\sqrt {a^2-c^2}=\frac{4\sqrt 3}{3}$. Thus the length of the minor axes is $2b=\frac{8\sqrt 3}{3}$
