Answer
$r=\frac{12}{5+3\sin\theta}$
Work Step by Step
Find the directrix:
$r=4\csc\theta$
$r=\frac{4}{\sin\theta}$
$r\sin\theta=4$
$y=4$
Using Theorem 6 with $e=0.6=\frac{3}{5}$ and $d=4$, and using part (c) of Figure 2, the ellipse with the focus at the origin, the eccentricity $0.6$, and the directrix $r=4\csc\theta$ has the following the polar equation:
$r=\frac{\frac{3}{5}\cdot 4}{1+\frac{3}{5}\sin\theta}$
$r=\frac{\frac{12}{5}}{1+\frac{3}{5}\sin\theta}$
$r=\frac{12}{5+3\sin\theta}$
