Calculus (3rd Edition)

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