Answer
$\frac{x^2}{16}-\frac{y^2}{64}=1$
See graph.
Work Step by Step
1. Given vertices $(-4,0)$ and $(4,0)$, asymptote $y=2x$, we have a horizontal transverse axis, center $(0,0)$ (midpoint of vertices), $a=4, \frac{b}{a}=2$ and $b=8, c=\sqrt {a^2+b^2}=\sqrt {16+64}=4\sqrt 5$,
2. Thus the equation is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ or $\frac{x^2}{16}-\frac{y^2}{64}=1$
3. See graph.
