Answer
(a) See graph and explanations.
(b) directrix $y=-3$. vertices $V_1(6, \frac{\pi}{2})$ and $V_2(\frac{6}{5},\frac{3\pi}{2})$
(c) center $C(\frac{9}{5}, \frac{\pi}{2})$. length of the major axes $\frac{36}{5}$. length of the minor axes $\frac{18\sqrt 3}{5}$
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 $r=\frac{2}{1-\frac{2}{3}sin\theta}$, we can identify that $e=\frac{2}{3}\lt1$ 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 below of the curve.
Step 3. See graph.
(b) With $e=\frac{2}{3}$ and $ed=2$, we have $d=3$, we can identify that the directrix as $y=-d=-3$. To find the vertices, let $\theta=\pi/2, 3\pi/2$, we have $r(\pi/2)=6$ and $r(3\pi/2)=\frac{6}{5}$, thus we have the vertices $V_1(6, \frac{\pi}{2})$ and $V_2(\frac{6}{5},\frac{3\pi}{2})$ as indicated in the figure.
(c) The center of the ellipse is half way between the two vertices, thus we have $r(c)=6/2-\frac{6}{5}=\frac{9}{5}$ and the center is at $C(\frac{9}{5}, \frac{\pi}{2})$.
The length of the major axes is the distance between the two vertices, thus $2a=6+\frac{6}{5}=\frac{36}{5}$.
With $a=\frac{18}{5}, c=\frac{9}{5}$ (origin is one of the focus), we have $b=\sqrt {a^2-c^2}=\frac{9\sqrt 3}{5}$. Thus the length of the minor axes is $2b=\frac{18\sqrt 3}{5}$
