Consider a hyperbola H : x ^{2}-2 y ^{2}=4. L | vedclass

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Question

$\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$

$e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{\frac{3}{2}}$

$	herefore$ Focus $F ( ae , 0) \Rightarrow F (\sqrt{6},

Vedclass · Accepted Answer

$\frac{7}{\sqrt{6}}-2$

Vedclass · Answer

$\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$

$e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{\frac{3}{2}}$

$	herefore$ Focus $F ( ae , 0) \Rightarrow F (\sqrt{6}, 0)$

equation of tangent at $P$ to the hyperbola is

$2 x-y \sqrt{6}=2$

tangent meet $x$ -axis at $Q(1,0)$

And latus rectum $x=\sqrt{6}$ at $R\left(\sqrt{6}, \frac{2}{\sqrt{6}}(\sqrt{6}-1)ight)$

$	herefore$ Area of $\Delta_{ QFR }=\frac{1}{2}(\sqrt{6}-1) \cdot \frac{2}{\sqrt{6}}(\sqrt{6}-1)$

$=\frac{7}{\sqrt{6}}-2$

Consider a hyperbola $H : x ^{2}-2 y ^{2}=4$. Let the tangent at a point $P (4, \sqrt{6})$ meet the $x$ -axis at $Q$ and latus rectum at $R \left( x _{1}, y _{1}\right), x _{1}>0 .$ If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\Delta QFR$ is equal to ....... .

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