Question

Find the equation of the hyperbola whose foci are $(0,\pm \sqrt{10})$ and passing through the point $(2,3)$.

Answer 1

Since, the foci of the given hyperbola are $F^{\prime }(0,-\sqrt{10})$ and $F(0,\sqrt{10})$ which lie on the X-axis and mid-point of the segment $F^{\prime }F$ is $(0,0)$. Therefore origin is the centre of the hyperbola and its transverse axis lies along Y-axis, hence the equation of the hyperbola can be taken as
$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ .........(i)

As the foci are $(0,\pm \sqrt{10})$
So, $ae=\sqrt{10}$

We know that
$b^2=a^2{e}^2-a^2$
$\Rightarrow b^2=10-a^2$ .........(ii)

Substituting this value of $b^2$ in (i), we get
$\frac{y^2}{a^2}-\frac{x^2}{10-a^2}=1$

Since the hyperbola passes through the point $(2,3)$, we get
$\frac{9}{a^2}-\frac{4}{10-a^2}=1$
$9(10-a^2)-4a^2=a^2(10-a^2)$
$\Rightarrow 90-13a^2=10a^2-a^4$
$\Rightarrow a^4-23a^2+90=0$
$\Rightarrow (a^2-5)(a^2-18)=0$
$\Rightarrow a^2=5,18$

If $a^2=5,b^2=10-a^2=10-5=5$

If $a^2=18,b^2=10-18=-8$ (not possible)

So, we get $a^2=5,b^{=}5$

Hence from (i), the equation of the hyperbola is $\frac{y^2}{5}-\frac{x^2}{5}=1$
i.e., $y^2-x^2=5$.