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A hyperbola passes through the point $P\left( {\sqrt 2 ,\sqrt 3 } \right)$ has foci at $\left( { \pm 2,0} \right)$. Then the tangent to this hyperbola at $P$ also passes through the point
$\left( { - \sqrt 2 , - \sqrt 3 } \right)$
$\left( {3\sqrt 2 ,2\sqrt 3 } \right)$
$\left( {2\sqrt 2 ,3\sqrt 3 } \right)$
$\left( {3,\sqrt 2 } \right)$
Solution
Equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
foci is $(\pm 2,0) \Rightarrow a c=2 \Rightarrow a^{2} c^{2}=4$
Since $b^{2}=a^{2}\left(e^{2}-1\right)$
$b^{2}=a^{2} e^{2}-a^{2}$
$\therefore \quad a^{2}+b^{2}=4$ …….$(1)$
Hyperbola passes through $(\sqrt{2}, \sqrt{3})$
$\therefore \frac{2}{a^{2}}-\frac{3}{b^{2}}=1$ ……..$(2)$
$\frac{2}{4-b^{2}} \frac{-3}{b^{2}}=1$
$\Rightarrow b^{4}+b^{2}-12=0$
$\Rightarrow\left(b^{2}-3\right)\left(b^{2}+4\right)=0$
$\Rightarrow b^{2}=3$
$b^{2}=-4 \quad$ (Not possible)
For $b^{2}=3$
$\Rightarrow a^{2}=1$
$\therefore \frac{x^{2}}{1}-\frac{y^{2}}{3}=1$
Equation of tangent is $\frac{\sqrt{2} x}{1}-\frac{\sqrt{3} y}{3}=1$
Clearly $(2 \sqrt{2}, 3 \sqrt{3})$ satisfies it.