QJEEAdv08Mafaba: JEE Advanced 2008 Mathematics - Class 11 JEEAdv08Mafaba Solution: Consider a branch of the hyperbola with vertex at the point .…

We have,

${x^2} - 2{y^2} - 2\sqrt 2 x - 4\sqrt 2 y - 6 = 0$

${{{{(x - \sqrt 2 )}^2}} \over 4} - {{{{(y + \sqrt 2 )}^2}} \over 2} = 1$

$a = 2,b = \sqrt 2$

$\Rightarrow e = \sqrt {{3 \over 2}}$

Now, $e = \sqrt {{{{a^2} + {b^2}} \over {{a^2}}}} = \sqrt {{3 \over 2}}$

Area, $= {1 \over 2}$ $\times$ Base $\times$ Height

$= {1 \over 2}a(e - 1){{{b^2}} \over a}$

$= {1 \over 2}{{(\sqrt 3 - \sqrt 2 ) \times 2} \over {\sqrt 2 }}$

$= {{\sqrt 3 - \sqrt 2 } \over {\sqrt 2 }}$

$= \left( {\sqrt {{3 \over 2}} - 1} \right)$