Answer
$\dfrac{y^2}{9}-\dfrac{(x-2)^2}{9/4}=1$
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 : $(2,0)$
We have: $ a=3$
$\dfrac{y^2}{9}-\dfrac{(x-2)^2}{b^2}=1$
or, $\dfrac{25}{9}-\dfrac{4}{b^2}=1$
$\implies b=\dfrac{3}{2}$
Now, $\dfrac{y^2}{9}-\dfrac{(x-2)^2}{9/4}=1$
