Answer
Equation of the conic section:
${\left( {\frac{x}{{64}}} \right)^2} + {\left( {\frac{y}{{24\sqrt 7 }}} \right)^2} = 1$
Work Step by Step
We have $c=8$ and $e = \frac{1}{8}$. By Theorem 4 of Section 12.5: $e = \frac{c}{a}$. So,
$\frac{1}{8} = \frac{8}{a}$, ${\ \ \ }$ $a=64$.
Since the foci are located at the $x$-axis, $\left( { \pm 64,0} \right)$ are the focal vertices. By Theorem 1 of Section 12.5, this is the equation of an ellipse in standard position, where $a>b$. So,
$b = \sqrt {{a^2} - {c^2}} = \sqrt {{{64}^2} - {8^2}} = 24\sqrt 7 $
Substituting $a$ and $b$ in Eq. (5) of Theorem 1 (Section 12.5) gives
${\left( {\frac{x}{{64}}} \right)^2} + {\left( {\frac{y}{{24\sqrt 7 }}} \right)^2} = 1$