Hyperbola 1 Question 9
10. 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 latusrectum. If $C$ is the focus of the hyperbola nearest to the point $A$, then the area of the $\triangle A B C$ is
(a) $1-\sqrt{2 / 3}$ sq unit
(2008, 3M)
(b) $\sqrt{3 / 2}-1$ sq unit
(c) $1+\sqrt{2 / 3}$ sq unit
(d) $\sqrt{3 / 2}+1$ sq unit
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Answer:
Correct Answer: 10. (d)
Solution:
- Given equation can be rewritten as focal chord
$$ \frac{(x-\sqrt{2})^{2}}{4}-\frac{(y+\sqrt{2})^{2}}{2}=1 $$
For point $A(x, y), e=\sqrt{1+\frac{2}{4}}=\sqrt{\frac{3}{2}}$
$$ \begin{aligned} \Rightarrow & & x-\sqrt{2} & =2 \\ \Rightarrow & & x & =2+\sqrt{2} \end{aligned} $$
For point $C(x, y), x-\sqrt{2}=a e=\sqrt{6} \Rightarrow x=\sqrt{6}+\sqrt{2}$
Now, $\quad A C=\sqrt{6}+\sqrt{2}-2-\sqrt{2}=\sqrt{6}-2$
and $\quad B C=\frac{b^{2}}{a}=\frac{2}{2}=1$
$\therefore$ Area of $\triangle A B C=\frac{1}{2} \times(\sqrt{6}-2) \times 1=\sqrt{\frac{3}{2}}-1$ sq unit
