ExamBro
ExamBro
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

  1. A
    \(1-\sqrt{\frac{2}{3}}\) sq unit
  2. B
    \(\sqrt{\frac{3}{2}}-1\) sq unit
  3. C
    \(1+\sqrt{\frac{2}{3}}\) sq unit
  4. D
    \(\sqrt{\frac{3}{2}}+1\) sq unit
Verified Solution

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 \(A(x, y)\),


\[
\begin{aligned}
& e=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}} \\
& \therefore x-\sqrt{2}=2 \Rightarrow x=2+\sqrt{2} \\
& \text { For } C(x, y), x-\sqrt{2}=a e=\sqrt{6} \\
& \Rightarrow \quad x=\sqrt{6}+\sqrt{2} \\
& \text { Now, } A C=\sqrt{6}+\sqrt{2}-2-\sqrt{2}=\sqrt{6}-2 \\
& \qquad B C=\frac{b^2}{a}=\frac{2}{2}=1 \\
& \text { Area of } \Delta A B C=\frac{1}{2} \times(\sqrt{6}-2) \times B C \\
& =\frac{1}{2} \times(\sqrt{6}-2) \times 1=\sqrt{\frac{3}{2}}-1
\end{aligned}
\]
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app