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Let $\mathrm{S}$ be the focus of the hyperbola $x^2-2 y^2=1$ lying on the positive $\mathrm{X}$-axis. Let $\mathrm{P}(-1,1)$ be a given point. Then the area of the triangle formed by the line PS with the coordinate axes is (in sq. units)
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The correct answer is:
$\frac{3}{2(2+\sqrt{6})}$
Given hyperbola $x^2-2 y^2=1$ can be written as
$$
\Rightarrow \mathrm{a}=1, \mathrm{~b}=\frac{1}{\sqrt{2}} \text { hence, } \mathrm{e}=\sqrt{\frac{3}{2}}
$$
Now from fig. Area made by co-ordinate axes and the line $\overline{P S}$ will be
Area of $\Delta \mathrm{OSB}=\frac{1}{2} \mathrm{OS} \times \mathrm{OB}$
$$
=\frac{1}{2} \times \frac{3}{(2+\sqrt{6})}
$$
$$
\Rightarrow \mathrm{a}=1, \mathrm{~b}=\frac{1}{\sqrt{2}} \text { hence, } \mathrm{e}=\sqrt{\frac{3}{2}}
$$
Now from fig. Area made by co-ordinate axes and the line $\overline{P S}$ will be
Area of $\Delta \mathrm{OSB}=\frac{1}{2} \mathrm{OS} \times \mathrm{OB}$
$$
=\frac{1}{2} \times \frac{3}{(2+\sqrt{6})}
$$
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