Answer
Equation of the conic section:
${\left( {\frac{x}{8}} \right)^2} + {\left( {\frac{y}{{\sqrt {61} }}} \right)^2} = 1$
Work Step by Step
We have $a=8$ and $c = \sqrt 3 $. Since the foci are located at the $x$-axis, $\left( { \pm 8,0} \right)$ 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 - 3} = \sqrt {61} $
Substituting $a$ and $b$ in Eq. (5) of Theorem 1 (Section 12.5) gives
${\left( {\frac{x}{8}} \right)^2} + {\left( {\frac{y}{{\sqrt {61} }}} \right)^2} = 1$
