Find the equation of the hyperbola satisfying the give conditions: Foci

Solution:

Foci $(0, \pm \sqrt{10})$, 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 $(0, \pm \sqrt{10}), c=\sqrt{10}$.

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 $(2,3)$,

$\frac{9}{a^{2}}-\frac{4}{b^{2}}=1$ $\ldots(2)$

From equations (1) and (2), we obtain

$\frac{9}{a^{2}}-\frac{4}{\left(10-a^{2}\right)}=1$

$\Rightarrow 9\left(10-a^{2}\right)-4 a^{2}=a^{2}\left(10-a^{2}\right)$

$\Rightarrow 90-9 a^{2}-4 a^{2}=10 a^{2}-a^{4}$

$\Rightarrow a^{4}-23 a^{2}+90=0$

$\Rightarrow a^{4}-18 a^{2}-5 a^{2}+90=0$

$\Rightarrow a^{2}\left(a^{2}-18\right)-5\left(a^{2}-18\right)=0$

$\Rightarrow\left(a^{2}-18\right)\left(a^{2}-5\right)=0$

$\Rightarrow a^{2}=18$ or 5

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 $\frac{y^{2}}{5}-\frac{x^{2}}{5}=1$.