Find the equation of the hyperbola satisfying the give conditions: Foci $\left(0,\pm \sqrt{10}\right)$
passing through $(2,3)$

Here the foci are on the $y$-axis.
Therefore, the equation of the hyperbola is of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
Since the foci are $\left(0,\pm \sqrt{10}\right)\Rightarrow ae=c=\sqrt{10}$
We know that $a^2+b^2=c^2$
$\therefore$ $a^2+b^2=10$
$\Rightarrow b^2=10-a^2...(1)$
Since the hyperbola passes through point $(2,3)$
$\frac{9}{a^2}-\frac{4}{b^2}=1\; ...(2)$
From equation (1) and (2), we obtain
$\frac{9}{a^2}-\frac{4}{\left(10-a^2\right)}=1$
$\Rightarrow 9\left(10-a^2\right)-4a^2=a^2\left(10-a^2\right)$
$\Rightarrow 90-9a^2-4a^2=10a^2-a^4$
$\Rightarrow a^4-23a^2+90=0$
$\Rightarrow$$a^4-18a^2-5a^2+90=0$
$\Rightarrow a^2\left(a^2-18\right)-5\left(a^2-18\right)=0$
$\Rightarrow$$\left(a^2-18\right)\left(a^2-5\right)=0$
$\Rightarrow$$a^2=18or5$
In hyperbola $c>a$, i.e., $c{}^2>a{}^2$
$\therefore a^2=5$
$\Rightarrow$$b^2=10-a^2=10-5=5$
Thus the equation of the hyperbola is $\frac{y^2}{5}-\frac{x^2}{5}=1$