Question

Consider a branch of the hyperbola $x^2 - 2y^2 - 2\sqrt2x - 4\sqrt2y - 6 = 0$ 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 $\Delta ABC$ is

• A. (a)$1 - \sqrt\frac{2}{3}$ sq unit
• B. (b)$\sqrt\frac{3}{2} - 1$ sq unit
• C. (c)$1 +\sqrt\frac{2}{3}$ sq unit
• D. (d)$\sqrt\frac{3}{2} + 1$sq unit

Answer 1

The Correct Option is B

Given equation can be rewritten as focal chord
$\frac{(x-\sqrt2)^2}{4} - \frac{(y+\sqrt2)^2}{2}=1$
For point A(x, y), e$=\sqrt {1+\frac{2}{4}}=\sqrt\frac{3}{2}$
$\Rightarrow x-\sqrt2=2\Rightarrow x=2+\sqrt2$
$x-\sqrt2=ae=\sqrt6 \Rightarrow x=\sqrt6+\sqrt2$
Now, AC$=\sqrt6+\sqrt2-2-\sqrt2=\sqrt6-2$
and BC$=\frac{b^2}{a}=\frac{2}{2}=1$
$\therefore Area of \Delta ABC=\frac{1}{2}\times(\sqrt 6-2)\times1=\sqrt\frac{3}{2}-1 sq unit$