Answer
$\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{64}=1$
Work Step by Step
The vertices (endpoints of the traverse axis) are on the x-axis,
so the traverse axis is horizontal.
Th$\mathrm{e}$ equation has form
$\displaystyle \frac{(x-h)^{2}}{a^{2}}- \displaystyle \frac{(y-k)^{2}}{b^{2}}=1$
From endpoint to endpoint of the traverse axis,
the distance is $8$ units.
$2a=8$
$a=4,\qquad(a^{2}=16)$
The center is their midpoint: $(0,0)=(h,k).$
The asymptotes (for horizontal traverse axis) are
$y=\displaystyle \frac{b}{a}x$ and $y=-\displaystyle \frac{b}{a}x$.
So, given the equation of one of them, $y=2x,$
$\displaystyle \frac{b}{a}=2\qquad/\times a$
$b=2a=2\cdot 4=8,\qquad (b^{2}=64)$
The equation is
$\displaystyle \frac{x^{2}}{16}-\frac{y^{2}}{64}=1$
