Answer
$\displaystyle \frac{y^{2}}{36}-\frac{x^{2}}{9}=1$
Work Step by Step
The vertices (endpoints of the traverse axis) are above each other,
so the traverse axis is vertical.
Th$e$ equation has form
$\displaystyle \frac{(y-k)^{2}}{a^{2}}- \displaystyle \frac{(x-h)^{2}}{b^{2}}=1$
From endpoint to endpoint of the traverse axis,
the distance is 12 units.
$2a=12$
$a=6,\qquad(a^{2}=36)$
The center is their midpoint: $(0,0)=(h,k).$
The asymptotes (for vertical traverse axis) are
$y=\displaystyle \frac{a}{b}x$ and $y=-\displaystyle \frac{a}{b}x$
So, given the equation of one of them, $y=2x,$
$\displaystyle \frac{a}{b}=2\qquad/\times\frac{b}{2}$
$\displaystyle \frac{a}{2}=b$
$b=\displaystyle \frac{6}{2}=3,\qquad (b^{2}=9)$
The equation is
$\displaystyle \frac{y^{2}}{36}-\frac{x^{2}}{9}=1$
