Answer
Graph V
Work Step by Step
Find $e$ and $d$:
$r=\frac{12}{4+3\sin\theta}$
$\frac{ed}{1+e\sin\theta}=\frac{\frac{12}{4}}{1+\frac{3}{4}\sin\theta}$
$e=\frac{3}{4}$ and $ed=\frac{12}{4}$
$e=\frac{3}{4}$ and $d=4$
Using Theorem 6, and using part (c) of Figure 2, the given polar equation represents a vertical ellipse with the eccentricity $\frac{3}{4}$ and the directrix $y=4$.
It is shown in Graph V.