Answer
(a) See graph and explanations.
(b) directrix $y=4$, vertices $V_1(\frac{12}{7}, \frac{\pi}{2})$ and $V_2(12, \frac{3\pi}{2})$
(c) center $C(\frac{36}{7}, \frac{3\pi}{2})$.
length of the major axes $\frac{96}{7}$.
length of the minor axes $\frac{24\sqrt {7}}{7}$
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 $r=\frac{3}{1+\frac{3}{4}sin\theta}$, we can identify that $e=\frac{3}{4}\lt1$ by comparing it with a standard equation in step-1. Thus the graph of the equation will be an ellipse with a directrix above the curve.
Step 3. See graph.
(b) With $e=\frac{3}{4}$ and $ed=3$, we have $d=4$, we can identify that the directrix as $y=d=4$. To find the vertices, let $\theta=\pi/2, 3\pi/2$, we have $r(\pi/2)=\frac{12}{7}$ and $r(3\pi/2)=12$, thus we have the vertices $V_1(\frac{12}{7}, \frac{\pi}{2})$ and $V_2(12, \frac{3\pi}{2})$ as indicated in the figure.
(c) The center of the ellipse is half way between the two vertices, thus we have $r(c)=12-(12+\frac{12}{7})/2=\frac{36}{7}$ and the center is at $C(\frac{36}{7}, \frac{3\pi}{2})$.
The length of the major axes is the distance between the two vertices, thus $2a=12+\frac{12}{7}=\frac{96}{7}$.
With $a=\frac{48}{7}, c=\frac{36}{7}$ (origin is one of the focus), we have $b=\sqrt {a^2-c^2}=\frac{12\sqrt {7}}{7}$. Thus the length of the minor axes is $2b=\frac{24\sqrt {7}}{7}$
