#### Answer

We use Theorem 6 to show that the ellipse $r = \frac{{de}}{{1 + e\cos \theta }}$ has eccentricity $e>0$ with focus at the origin and directrix $x=d$. Then obtain the following results:
$\begin{array}{*{20}{c}}
{Point}&{A'}&{{F_2}}&C&{{F_1}}&A\\
{x - coordinate}&{ - \frac{{de}}{{1 - e}}}&{ - \frac{{2d{e^2}}}{{1 - {e^2}}}}&{ - \frac{{d{e^2}}}{{1 - {e^2}}}}&0&{\frac{{de}}{{1 + e}}}
\end{array}$

#### Work Step by Step

According to Theorem 6, the ellipse $r = \frac{{de}}{{1 + e\cos \theta }}$ has eccentricity $e>0$ with focus at the origin and directrix $x=d$. This is depicted in Figure 24 where one of the foci is at ${F_1} = \left( {0,0} \right)$.
From Figure 24 we see that the points $A$ and $A'$ are the focal vertices of the ellipse, corresponding to $\theta=0$ and $\theta=\pi$, respectively. Substituting $\theta=0$ and $\theta=\pi$ in $r$ give the $x$-coordinates of $A$ and $A'$, denoted by ${x_A}$ and ${x_{A'}}$, respectively:
${x_A} = \frac{{de}}{{1 + e}}$, ${\ \ \ }$ ${x_{A'}} = - \frac{{de}}{{1 - e}}$
Note that for an ellipse, we have $0 \le e < 1$; and since $A'$ is on the negative $x$-axis, we assign a negative sign to ${x_{A'}}$.
From this result we obtain the $x$-coordinate of the center $C$:
${x_C} = \frac{{{x_A} + {x_{A'}}}}{2} = \frac{1}{2}\left( {\frac{{de}}{{1 + e}} - \frac{{de}}{{1 - e}}} \right)$
${x_C} = \frac{1}{2}\left( {\frac{{de - d{e^2} - de - d{e^2}}}{{1 - {e^2}}}} \right) = \frac{1}{2}\left( {\frac{{ - 2d{e^2}}}{{1 - {e^2}}}} \right)$
${x_C} = \frac{{ - d{e^2}}}{{1 - {e^2}}}$
Since ${F_1} = \left( {0,0} \right)$, the distance between ${F_1}$ and $C$ is $\frac{{d{e^2}}}{{1 - {e^2}}}$. Therefore, the $x$-coordinate of ${F_2}$ is ${x_{{F_2}}} = \frac{{ - 2d{e^2}}}{{1 - {e^2}}}$.
We summarize the results in the following table:
$\begin{array}{*{20}{c}}
{Point}&{A'}&{{F_2}}&C&{{F_1}}&A\\
{x - coordinate}&{ - \frac{{de}}{{1 - e}}}&{ - \frac{{2d{e^2}}}{{1 - {e^2}}}}&{ - \frac{{d{e^2}}}{{1 - {e^2}}}}&0&{\frac{{de}}{{1 + e}}}
\end{array}$