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  4. Consider a hyperbola H: x 2-2 y 2=4. Let the tangent at a point P (4, √6) meet the x -axis at Q and latus rectum at R(x1, y1), x1>0 . If F is a focus of H which is nearer to the point P, then the area of Δ QFR is equal to

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Q. 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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Solution:

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$\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$
$e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{\frac{3}{2}}$
$\therefore $ 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 $Q(1,0)$
$\&$ latus rectum $x=\sqrt{6}$ at $R\left(\sqrt{6}, \frac{2}{\sqrt{6}}(\sqrt{6}-1)\right)$
$\therefore $ Area of $\Delta_{ QFR }=\frac{1}{2}(\sqrt{6}-1) \cdot \frac{2}{\sqrt{6}}(\sqrt{6}-1)$
$=\frac{7}{\sqrt{6}}-2$