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Consider a hyperbola  \mathrm{H: x^2-2 y^2=4.}Let the tangent at a point \mathrm{P(4, \sqrt{6})} meet the x-axis at Q and latus rectum at \mathrm{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 \triangle Q F R is equal to
 

Option: 1

4 \sqrt{6}


Option: 2

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


Option: 3

\sqrt{6}-1


Option: 4

4 \sqrt{6}-1


Answers (1)

best_answer

We have, \mathrm{x^2-2 y^2=4 \Rightarrow \frac{x^2}{4}-\frac{y^2}{2}=1}
\mathrm{ \Rightarrow e=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}}}

Thus, foci =\mathrm{ ( \pm \sqrt{6}, 0)} and \mathrm{ F=(\sqrt{6}, 0)}

Equation of tangent at point \mathrm{ P(4, \sqrt{6}) ~is~ 4 x-2 \sqrt{6} y=4} which meet x-axis at

\mathrm{ Q(1,0)} and latus rectum at
\mathrm{R\left(\sqrt{6}, \frac{4 \sqrt{6}-4}{2 \sqrt{6}}\right) ~or ~R\left(\sqrt{6}, \frac{4(\sqrt{6}-1)}{2 \sqrt{6}}\right)}
\mathrm{\therefore ~Area ~of~ \triangle Q F R=\frac{1}{2}\left[(\sqrt{6}-1) \times \frac{4(\sqrt{6}-1)}{2 \sqrt{6}}\right]=\frac{7}{\sqrt{6}}-2}

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Nehul

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