Chapter 10 - Section 10.6 - Conic Sections in Polar - 10.6 Exercises - Page 717: 2

$r=\frac{6}{3+\sin\theta}$

Using Theorem with $e=\frac{1}{3}$ and $d=6$, and using part (c) of Figure 2, the ellipse with the focus at the origin, the eccentricity $\frac{1}{3}$ and the directrix $y=6$ has the following equation $r=\frac{\frac{1}{3}\cdot 6}{1+\frac{1}{3}\sin\theta}$ $r=\frac{2}{1+\frac{1}{3}\sin\theta}$ $r=\frac{6}{3+\sin\theta}$