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

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(C) The given hyperbola is $x^{2}-2y^{2}=4$,which can be written as $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$. Here,$a^{2}=4$ and $b^{2}=2$. The eccentricit

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

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(C) The given hyperbola is $x^{2}-2y^{2}=4$,which can be written as $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$. Here,$a^{2}=4$ and $b^{2}=2$. The eccentricity $e$ is given by $e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}}$. The focus $F$ is $(ae, 0) = (2 \cdot \sqrt{\frac{3}{2}}, 0) = (\sqrt{6}, 0)$. The equation of the tangent at $P(4, \sqrt{6})$ to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ is $\frac{x x_{1}}{4}-\frac{y y_{1}}{2}=1$. Substituting $(x_{1}, y_{1}) = (4, \sqrt{6})$,we get $\frac{4x}{4}-\frac{\sqrt{6}y}{2}=1$,which simplifies to $x - \frac{\sqrt{6}}{2}y = 1$,or $2x - y\sqrt{6} = 2$. To find $Q$,set $y=0$ in the tangent equation: $2x=2 \Rightarrow x=1$. So,$Q(1, 0)$. To find $R$,substitute $x=\sqrt{6}$ (the latus rectum) into the tangent equation: $2(\sqrt{6}) - y\sqrt{6} = 2 \Rightarrow y\sqrt{6} = 2\sqrt{6}-2 \Rightarrow y = 2 - \frac{2}{\sqrt{6}}$. Thus,$R(\sqrt{6}, 2 - \frac{2}{\sqrt{6}})$. The area of $\Delta QFR$ with vertices $Q(1, 0)$,$F(\sqrt{6}, 0)$,and $R(\sqrt{6}, 2 - \frac{2}{\sqrt{6}})$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Base $QF = |\sqrt{6}-1|$. Height $FR = |2 - \frac{2}{\sqrt{6}}|$. Area $= \frac{1}{2} 	imes (\sqrt{6}-1) 	imes 2(1 - \frac{1}{\sqrt{6}}) = (\sqrt{6}-1) 	imes \frac{\sqrt{6}-1}{\sqrt{6}} = \frac{(\sqrt{6}-1)^{2}}{\sqrt{6}} = \frac{6+1-2\sqrt{6}}{\sqrt{6}} = \frac{7}{\sqrt{6}}-2$.

Consider a hyperbola $H : x^{2}-2y^{2}=4$. Let the tangent at a point $P(4, \sqrt{6})$ meet the $x$-axis at $Q$ and the latus rectum at $R(x_{1}, y_{1})$,where $x_{1}>0$. 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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