एक अविपरवलय बिंदु P(sqrt{2}, sqrt{3}) से होकर | vedclass

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Question

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

Vedclass · Accepted Answer

$\left( {2\sqrt 2 ,3\sqrt 3 } \right)$

Vedclass · Answer

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 $b^{2}=a^{2}\left(e^{2}-1ight)$

$b^{2}=a^{2} e^{2}-a^{2}$

$	herefore \quad a^{2}+b^{2}=4$            .......$(1)$

Hyperbola passes through $(\sqrt{2}, \sqrt{3})$

$	herefore \frac{2}{a^{2}}-\frac{3}{b^{2}}=1$          ........$(2)$

$\frac{2}{4-b^{2}} \frac{-3}{b^{2}}=1$

$\Rightarrow b^{4}+b^{2}-12=0$

$\Rightarrow\left(b^{2}-3ight)\left(b^{2}+4ight)=0$

$\Rightarrow b^{2}=3$

$b^{2}=-4 \quad$ (Not possible)

For $b^{2}=3$

$\Rightarrow a^{2}=1$

$	herefore \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.

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