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