Answer
Equation of the ellipse in rectangular coordinates:
${\left( {\frac{{x + 3}}{5}} \right)^2} + {\left( {\frac{y}{4}} \right)^2} = 1$
Work Step by Step
Write
$r = \frac{{16}}{{5 + 3\cos \theta }}$
$r = \frac{{16/5}}{{1 + \left( {3/5} \right)\cos \theta }} = \frac{{\left( {3/5} \right)\left( {16/3} \right)}}{{1 + \left( {3/5} \right)\cos \theta }}$
By Eq. (11) of Theorem 6, this is the polar equation of a conic section, with eccentricity $e = \frac{3}{5}$, a focus at the origin and directrix $x = \frac{{16}}{3}$. Since $0 \le e < 1$, by Theorem 4, this is an ellipse. This implies that $x$-axis is the focal axis.
Using the results of Exercise 60, we have
$\begin{array}{*{20}{c}}
{Point}&{A'}&{{F_2}}&C&{{F_1}}&A\\
{x - coordinate}&{ - \frac{{de}}{{1 - e}}}&{ - \frac{{2d{e^2}}}{{1 - {e^2}}}}&{ - \frac{{d{e^2}}}{{1 - {e^2}}}}&0&{\frac{{de}}{{1 + e}}}
\end{array}$
where $A$ and $A'$ are the focal vertices, ${F_1}$ and ${F_2}$ are the foci, $C$ is the center of the ellipse.
Substituting $e$ and $d$ in the equations for $x$-coordinates we get
$\begin{array}{*{20}{c}}
{Point}&{A'}&{{F_2}}&C&{{F_1}}&A\\
{x - coordinate}&{ - 8}&{ - 6}&{ - 3}&0&2
\end{array}$
From these results we have
1. Center $C=\left(-3,0\right)$
2. Focal vertices at $\left(-8,0\right)$ and $\left(2,0\right)$
3. Foci at $\left(-6,0\right)$ and $\left(0,0\right)$
Since the focal vertices are $\left(-8,0\right)$ and $\left(2,0\right)$, the semimajor axis is $10$.
The ellipse centered at the origin would have equation ${\left( {\frac{x}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$.
If the center is translated to $C=\left(-3,0\right)$, the equation has the form:
(1) ${\ \ \ }$ ${\left( {\frac{{x + 3}}{a}} \right)^2} + {\left( {\frac{y}{b}} \right)^2} = 1$
Notice that $a$ and $b$ are unchanged after translation in this equation. So, we will work with the ellipse as if it is in the standard position to obtain $a$ and $b$. Then, we substitute the results in equation (1) to obtain the equation of the ellipse when it is centered at $C=\left(-3,0\right)$.
Since the semimajor axis is $10$, in standard position the focal vertices are at $\left( { \pm a,0} \right) = \left( { \pm 5,0} \right)$. By Theorem 4, the eccentricity for an ellipse is $e = \frac{c}{a}$.
So,
$e = \frac{3}{5} = \frac{c}{5}$, ${\ \ \ }$ $c=3$
$b = \sqrt {{a^2} - {c^2}} = \sqrt {{5^2} - {3^2}} = \sqrt {16} = 4$
Substituting $a$ and $b$ in equation (1) gives
${\left( {\frac{{x + 3}}{5}} \right)^2} + {\left( {\frac{y}{4}} \right)^2} = 1$
