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Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$, then the area of the $\triangle A B C$ is
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The correct answer is:
$\sqrt{\frac{3}{2}}-1$ sq unit
$\sqrt{\frac{3}{2}}-1$ sq unit
The given equation can be rewritten as $\frac{(x-\sqrt{2})^2}{1}-\frac{(y+\sqrt{2})^2}{2}=1$ for $A(x, y)$,
$$
\begin{aligned}
& e=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}} \\
& \therefore x-\sqrt{2}=2 \Rightarrow x=2+\sqrt{2} \\
& \text { For } C(x, y), x-\sqrt{2}=a e=\sqrt{6} \\
& \Rightarrow \quad x=\sqrt{6}+\sqrt{2} \\
& \text { Now, } A C=\sqrt{6}+\sqrt{2}-2-\sqrt{2}=\sqrt{6}-2 \\
& \qquad B C=\frac{b^2}{a}=\frac{2}{2}=1 \\
& \text { Area of } \Delta A B C=\frac{1}{2} \times(\sqrt{6}-2) \times B C \\
& =\frac{1}{2} \times(\sqrt{6}-2) \times 1=\sqrt{\frac{3}{2}}-1
\end{aligned}
$$
$$
\begin{aligned}
& e=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}} \\
& \therefore x-\sqrt{2}=2 \Rightarrow x=2+\sqrt{2} \\
& \text { For } C(x, y), x-\sqrt{2}=a e=\sqrt{6} \\
& \Rightarrow \quad x=\sqrt{6}+\sqrt{2} \\
& \text { Now, } A C=\sqrt{6}+\sqrt{2}-2-\sqrt{2}=\sqrt{6}-2 \\
& \qquad B C=\frac{b^2}{a}=\frac{2}{2}=1 \\
& \text { Area of } \Delta A B C=\frac{1}{2} \times(\sqrt{6}-2) \times B C \\
& =\frac{1}{2} \times(\sqrt{6}-2) \times 1=\sqrt{\frac{3}{2}}-1
\end{aligned}
$$
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