Chapter 10 - Section 10.6 - Conic Sections in Polar - 10.6 Exercises - Page 717: 4

$r=\frac{15}{2-5\cos\theta}$

Using Theorem 6 with $e=\frac{5}{2}$ and $d=3$, and using part (b) of Figure 2, the hyperbola with the focus at the origin, the eccentricity $\frac{5}{2}$, and the directrix $x=-3$ has the following equation: $r=\frac{\frac{5}{2}\cdot 3}{1-\frac{5}{2}\cos\theta}$ $r=\frac{\frac{15}{2}}{1-\frac{5}{2}\cos\theta}$ $r=\frac{15}{2-5\cos\theta}$