Answer
$\dfrac{(y-1)^2}{1}-\dfrac{x^2}{3}=1$
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: $(0,1)$
We have: $c=2; a=1$
$b^2=a^2-c^2=2^2-1^2=\sqrt 3$
$\dfrac{(y-1)^2}{1^2}-\dfrac{x^2}{3}=1$
or, $\dfrac{(y-1)^2}{1}-\dfrac{x^2}{3}=1$
