Answer
$\dfrac{(x-3)^2}{4}-\dfrac{(y-2)^2}{5}=1$
Work Step by Step
The standard form of the equation of the hyperbola with a horizontal transverse axis can be expressed as: $\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
and 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: $\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
and vertices and foci have the form $(0, \pm, a) $ and $(0, \pm c)$.
The center is the midpoint of the vertices : $(3,2)$
We have: $c=3; a=2$
$b^2=c^2-a^2=3^2-2^2= 5$
$\dfrac{(x-3)^2}{2^2}-\dfrac{(y-2)^2}{5}=1$
or, $\dfrac{(x-3)^2}{4}-\dfrac{(y-2)^2}{5}=1$
