Question

Find the equation of hyperbola when, foci $(0,\pm \sqrt{10})$, passing through $(2,3)$

Answer 1

Consider the problem

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 $\left(0,\pm \sqrt{10}\right),c=\sqrt{10}$

we know that $a^2+b^2=c^2$

Therefore,

$\begin{array}{c}{a}^2+b^2=10\\ \Rightarrow b^2=10-a^2......\left(1\right)\end{array}$

Since the hyperbola passes through point $(2,3)$

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

Now, from equation $\left(1\right)$ and $\left(2\right)$

$\begin{array}{c}\frac{9}{a^2}-\frac{4}{\left(10-a^2\right)}=1\\ \Rightarrow 9\left(10-a^2\right)-4a^2=a^2\left(10-a^2\right)\\ \Rightarrow 90-9a^2-4a^2=10a^2-a^4\\ \Rightarrow a^4-23a^2+90=0\\ \Rightarrow a^4-18a^2-5a^2+90=0\\ \Rightarrow a^2\left(a^2-18\right)-5\left(a^2-18\right)=0\\ \Rightarrow \left(a^2-18\right)\left(a^2-5\right)=0\\ \Rightarrow a^2=18or5\end{array}$

In hyperbola, $c>a,i.e.,c^2>a^2$

Therefore, $a^2=5$

$\begin{array}{c}\Rightarrow b^2=10-a^2\\ =10-5=5\end{array}$

Hence, the equation of hyperbola is $\frac{y^2}{5}-\frac{x^2}{5}=1$