Answer
(a) $e=\frac{2}{3}$, ellipse.
(b) See graph.
Work Step by Step
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 $e=1$, an ellipse if $e\lt1$, and a hyperbola if $e\gt1$
(a) Rewrite the equation given in the Exercise as $r=\frac{10/3}{1-\frac{2}{3}sin\theta}$, we can identify that $e=\frac{2}{3}\lt1$ by comparing it with a standard equation above. Thus the graph of the equation will be an ellipse.
(b) See graph. Vertices $V_1(10, \frac{\pi}{2})$ and $V_2(2, \frac{3\pi}{2})$
