Question

Referred to the principle axes as the coordinate axes, find the equation of the hyperbola whose Foci are at $(0,\pm \sqrt{10})$ and which passes through the point $(2,3)$.

Answer 1

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

$t=5,18$

$\therefore a^2=t$

$\Rightarrow t=5$

$\therefore a^2=5$

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

$\Rightarrow b^2=10-5$

$\Rightarrow b^2=5$

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

$\therefore \frac{y^2}{a^2}-\frac{x^2}{b^2}=2$

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

Equation of hyperbola:

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

foci is on the $y-$axis;

$\therefore$ foci:- $(0,\pm c)=(0,\pm \sqrt{10})$

$\Rightarrow c=\sqrt{10}$

$c^2=a^2+b^2$

$\Rightarrow 10=a^2+b^2$

$\Rightarrow b^2=10-a^2....\left(1\right)$

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

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

Put $b^2=10-a^2$ (from equation $\left(1\right)$)

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

$\Rightarrow 90-9a^2-4a^2=a^2(10-a^2)$

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

let $a^2=t$