Answer
Vertex 1: $\theta=0$, $x = \frac{{ed}}{{e + 1}}$
Vertex 2: $\theta=\pi$, $x = \frac{{ed}}{{e - 1}}$
Work Step by Step
By Eq. (11) of Theorem 6, this is the polar equation of a conic section, with eccentricity $e$, a focus at the origin and directrix $x=d$. This implies that $x$-axis is the focal axis. So, the vertices correspond to $\theta=0$ and $\theta=\pi$ are
Vertex 1: $\theta=0$, $x = \left| {\frac{{de}}{{1 + e}}} \right| = \frac{{ed}}{{e + 1}}$
Vertex 2: $\theta=\pi$, $x = \left| { - \frac{{de}}{{1 - e}}} \right| = \frac{{ed}}{{e - 1}}$
