Question

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

Answer 1

Step $1$: Assuming equation of hyperbola
Given:
Foci are $(0,\pm \sqrt{10})$and the hyperbola is passing through $(2,3)$
As we can see that foci are on the $y$ − axis.
So, the equation of the hyperbola will be of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ $...\left(1\right)$

Step $2$: Finding the values of $a^2$ and $b^2$
Foci $=(0,\pm \sqrt{10})=(0,\pm c)$ i.e., $c=\sqrt{10}$
As we know that $c^2=a^2+b^2$
So, $(\sqrt{10}{)}^2=a^2+b^2$
$\Rightarrow a^2+b^2=10$
$\Rightarrow b^2=10-a^2$ $...\left(2\right)$

Equation $\left(1\right)$ passing through $(2,3)$
$\Rightarrow \frac{y^2}{a^2}-\frac{x^2}{b^2}$=1

$\Rightarrow \frac{3^2}{a^2}-\frac{2^2}{b^2}=1$

$\Rightarrow \frac{9}{a^2}-\frac{4}{b^2}=1$ $....\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$

$\Rightarrow \frac{9}{a^2}-\frac{4}{10-a^2}=1$

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

$\Rightarrow \frac{90-9a^2-4a^2}{a^2(10-a^2)}=1$

$\Rightarrow 90-13a^2=a^2(10-a^2)$

$\Rightarrow 90-13a^2=10a^2-a^4$

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

Let $a^2=x$

So, our equation becomes

$\Rightarrow x^2-23x+90=0$

$\Rightarrow x^2-18x-5x+90=0$

$\Rightarrow x(x-18)-5(x-18)=0$

$\Rightarrow (x-18)(x-5)=0$

$\Rightarrow x=5,18$

$\Rightarrow a^2=5,18$

Putting these values in equation $\left(2\right),$ we get

Case -$1$ : $a^2=5$

$\Rightarrow b^2=10-5=5$

Case-$2$ : $a^2=18$
$\Rightarrow b^2=10-18=-8$ (Not possible)

$\therefore a^2=5$ and $b^2=5$

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