JEE Advanced · Mathematics · 15. Hyperbola
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
- A
\(1-\sqrt{\frac{2}{3}}\) sq unit
- B
\(\sqrt{\frac{3}{2}}-1\) sq unit
- C
\(1+\sqrt{\frac{2}{3}}\) sq unit
- D
\(\sqrt{\frac{3}{2}}+1\) sq unit
Answer & Solution
Correct Answer
(B)
\(\sqrt{\frac{3}{2}}-1\) sq unit
Step-by-step Solution
Detailed explanation
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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