Answer
$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{32}=1$
Work Step by Step
The foci have equal y-coordinates,
so the transverse axis is horizontal
and its equation is of the form
$\displaystyle \frac{(y-k)^{2}}{a^{2}}- \displaystyle \frac{(x-h)^{2}}{b^{2}}=1$
The midpoint of the foci (the center):
$(\displaystyle \frac{0+0}{2},\frac{-6+6}{2})=(0,0)=(h,k)$
The vertices are a=2 units above/below the center.
$(a^{2}=4)$
From the foci, $c=6$.
The remaining thing to find is $b^{2}$:
$b^{2}=c^{2}-a^{2}=36-4=32$
The equation is
$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{32}=1$
