Answer
$\dfrac{x^2}{36} -\dfrac{(y-7)^2}{9}=1$
Work Step by Step
The standard form of the equation of the hyperbola with a horizontal transverse axis can be expressed as: $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
The vertices and foci have the form $(\pm a, 0) $ and $(\pm c,0)$.
The standard form of the equation of the hyperbola with a vertical
transverse axis can be expressed as: $\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
The vertices and foci have the form $(0, \pm, a) $ and $(0, \pm c)$.
$b/a=\dfrac{1}{2}$
So, $b=3$
$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$
or, $\dfrac{x^2}{36} -\dfrac{(y-7)^2}{9}=1$
