Answer
Graph IV
Work Step by Step
Find $e$ and $d$:
$r=\frac{5}{2+3\sin\theta}$
$\frac{ed}{1+\sin\theta}=\frac{\frac{5}{2}}{1+\frac{3}{2}\sin\theta}$
$e=\frac{3}{2}$ and $ed=\frac{5}{2}$
$e=\frac{3}{2}$ and $d=\frac{5}{3}$
Using Theorem 6, and using part (c) of Figure 2, the given polar equatin represents a vertical hyperbola with the eccentricity $\frac{3}{2}$ and the directrix $y=\frac{5}{3}$.
It is shown in Graph IV.
