#### Answer

Equation of the hyperbola:
${\left( {\frac{{y + 1}}{5}} \right)^2} - {\left( {\frac{x}{{\sqrt {39} }}} \right)^2} = 1$

#### Work Step by Step

From the vertices $\left(0,-6\right)$ and $\left(0,4\right)$ we obtain the center of the hyperbola at $C = \left( {0,\frac{{4 - 6}}{2}} \right) = \left( {0, - 1} \right)$.
Since the vertices are $\left(0,-6\right)$ and $\left(0,4\right)$, the distance between the vertices is $10$.
Since the focal vertices are on the y-axis, the hyperbola centered at the origin would have equation ${\left( {\frac{y}{a}} \right)^2} - {\left( {\frac{x}{b}} \right)^2} = 1$.
If the center is translated to $C=\left(0,-1\right)$, the equation has the form:
(1) ${\ \ \ }$ ${\left( {\frac{{y + 1}}{a}} \right)^2} - {\left( {\frac{x}{b}} \right)^2} = 1$
Notice that $a$ and $b$ are unchanged after translation in this equation. So, we will work with the hyperbola as if it is in the standard position to obtain $a$ and $b$. Then, we substitute the results in equation (1) to obtain the equation of the hyperbola when it is centered at $C=\left(0,-1\right)$.
Since the distance between vertices is $10$, in standard position the focal vertices are at $\left( {0, \pm a} \right) = \left( {0, \pm 5} \right)$. The foci $\left(0,-9\right)$ and $\left(0,7\right)$ in standard position is $\left( {0, \pm c} \right) = \left( {0, \pm \frac{{7 + 9}}{2}} \right) = \left( {0, \pm 8} \right)$.
By Theorem 2, we have $c = \sqrt {{a^2} + {b^2}} $. So,
$b = \sqrt {{c^2} - {a^2}} = \sqrt {{8^2} - {5^2}} = \sqrt {39} $
Substituting $a$ and $b$ in equation (1) gives
${\left( {\frac{{y + 1}}{5}} \right)^2} - {\left( {\frac{x}{{\sqrt {39} }}} \right)^2} = 1$