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The equation of the hyperbola in standard form whose foci are at $\mathrm{(0, \pm \sqrt{10})}$ and which passes through $\mathrm{(2,3)}$ is
Option: 1
$\mathrm{x^2-y^2=4}$

Option: 2
$\mathrm{x^2-y^2=5}$

Option: 3
$\mathrm{x^2-y^2=1}$

Option: 4
$\mathrm{x^2-y^2=-5}$

Answers (1)

best_answer

As the foci lie on the y-axis, the equation of hyperbola is

$\mathrm{ \frac{y^2}{b^2}-\frac{x^2}{a^2}=1 }$ $\mathrm{.....(i)}$

It passes through $(2,3).$ The condition is

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

For foci, be $\mathrm{=\sqrt{10} \Rightarrow b^2 e^2=10}$ $\mathrm{.....(iii)}$

Also $\mathrm{a^2=b^2\left(e^2-1\right)}$

Or $\mathrm{a^2=10-b^2}$ $\mathrm{.....(iv)}$

From (ii) and (iv),

$\mathrm{ \frac{9}{b^2}=1+\frac{4}{a^2}=1+\frac{4}{10-b^2} }$

$\mathrm{ \begin{array}{ll} \text { Or } & 90-9 b^2=14 b^2-b^4 \\\\ \text { Or } & b^4-23 b^2+90=0 \\\\ \text { Or } & \left(b^2-18\right)\left(b^2-5\right)=0 \\\\ \therefore & b^2=5,18 \end{array} }$
$\mathrm{\text { On using (iv) and } b^2=18 \text { gives } a^2=10-18=-8 \text {, which is not possible. }}$

$\mathrm{\therefore \quad b^2=5 \text { and then } a^2=10-5=5}$

Hence, the hyperbola is $\mathrm{x^2-y^2=5.}$

The answer is (b)

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