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

Solution: Here, Foci of hyperbola $=\left(\pm \sqrt{10},0\right)$
That means the transverse axis of the hyperbola is $Y$-axis.
So, the equation will be of the type,
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\to \left(1\right)$
Also, $c=\sqrt{10}$
In a hyperbola, $c^2=a^2+b^2$
Putting value of $c$,
$\Rightarrow {\left(\sqrt{10}\right)}^2=a^2+b^2$
$\Rightarrow b^2=10-a^2$
Putting vallue of $b^2$ in (1),
$\frac{y^2}{a^2}-\frac{x^2}{10}-a^2=1\to \left(2\right)$
As hyperbola is passing through $\left(2,3\right)$, $x=2$ and $y=3$ should satisfy (2).
$So,\frac{9}{a^2}-\frac{4}{10-a^2}=1$
$\Rightarrow 9\left(10-a^2\right)-4a^2=a^2\left(10-a^2\right)$
$\Rightarrow 90-9a^2-4a^2=10a^2-a^4$
$\Rightarrow a^4-23a^2+90=0$
$\Rightarrow a^4-18a^2-5a^2+90=0$
$\Rightarrow \left(a^2-18\right)\left(a^2-5\right)=0$
$\Rightarrow a^2=18\text{and}{a}^2=5$
But, $a^2$ can not be greater than $c^2$.
$\therefore a^2=5$
$b^2=10-5=5$
Putting values of $a^2$ and $b^2$ in (1),
$\frac{y^2}{5}-\frac{x^2}{5}=1$