Answer : $\large\frac{x^2}{5}-\frac{y^2}{5}\normalsize =-1$
Given data is the coordinates of foci are $(0,\pm\sqrt{10})$
The hyperbola passes through the point $(2,3)$.
Hence the equation of the hyperbola is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1$
It passes through $(2,3)$
$\therefore \large\frac{4}{a^2}-\frac{9}{b^2}$$=-1$
But $a^2=b^2(e^2-1)$
$\therefore \large\frac{4}{b^2(e2-1)}-\frac{9}{b^2}$$=-1$------(1)
The coordinates of the foci are $(0,\pm \sqrt{10})$
$\therefore be=\sqrt{10}$
$\Rightarrow b^2e^2=10$
Hence $\large\frac{4}{10-b^2}-\frac{9}{b^2}$$=-1$
On simplifying we get,
$4b^2-9(10-b^2)=-b^2(10-b^2)$
$\Rightarrow 4b^2-90+9b^2=-10b^2+b^4$
$\Rightarrow b^4-23b^2+90=0$
On factorizing we get,
$(b^2-18)(b^2-5)=0$
$\therefore b^2=18$ or $b^2=5$
$a^2=b^2(e^2-1)$
$a^2=10-b^2$
$\;\;\;\;=10-18=-8$ which is not possible.
$a^2=10-5=5$
Hence the required equation is $\large\frac{x^2}{5}-\frac{y^2}{5}$$=-1$