Answer
(a) See graph and explanations.
(b) directrix $x=6$. vertices $V_1(\frac{18}{7},0)$ and $V_2(18, \pi)$
(c) center $C(\frac{54}{7}, \pi)$. major axes $\frac{144}{7}$. minor axes $\frac{36\sqrt 7}{7}$.
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 $r=\frac{9/2}{1+\frac{3}{4}cos\theta}$, 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 to the right of the curve.
Step 3. See graph.
(b) With $e=\frac{3}{4}$ and $ed=9/2$, we have $d=6$, we can identify that the directrix as $x=d=6$.
To find the vertices, let $\theta=0, \pi$, we have $r(0)=\frac{18}{7}$ and $r(\pi)=18$, thus we have the vertices $V_1(\frac{18}{7},0)$ and $V_2(18, \pi)$ as indicated in the figure.
(c) The center of the ellipse is half way between the two vertices, thus we have $r(c)=18-(18+\frac{18}{7})/2=\frac{54}{7}$ and the center is at $C(\frac{54}{7}, \pi)$.
The length of the major axes is the distance between the two vertices, thus $2a=18+\frac{18}{7}=\frac{144}{7}$.
With $a=\frac{72}{7}, c=\frac{54}{7}$ (origin is one of the focus), we have $b=\sqrt {a^2-c^2}=\frac{18\sqrt 7}{7}$. Thus the length of the minor axes is $2b=\frac{36\sqrt 7}{7}$.
