Answer
Equation of the conic section:
${\left( {\frac{x}{8}} \right)^2} - {\left( {\frac{y}{6}} \right)^2} = 1$
Work Step by Step
We have $a=8$.
Since $y = \pm \frac{b}{a}x$, so $\frac{b}{a} = \frac{3}{4}$. It follows that $b=6$.
Since the vertices $\left( { \pm 8,0} \right)$ are located on the $x$-axis, the $x$-axis is the focal axis. By Theorem 2 of Section 12.5, this is a hyperbola in standard position. Substituting $a$ and $b$ in Eq. (7) of Theorem 2 (Section 12.5) we obtain the equation:
${\left( {\frac{x}{8}} \right)^2} - {\left( {\frac{y}{6}} \right)^2} = 1$
