JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 14)
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 $$ABC$$ is
$$1 - \sqrt {{2 \over 3}} $$
$$\sqrt {{3 \over 2}} - 1$$
$$1 + \sqrt {{2 \over 3}} $$
$$\sqrt {{3 \over 2}} + 1$$
Explanation
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
$$ = {1 \over 2}a(e - 1){{{b^2}} \over a}$$
$$ = {1 \over 2}{{(\sqrt 3 - \sqrt 2 ) \times 2} \over {\sqrt 2 }}$$
$$ = {{\sqrt 3 - \sqrt 2 } \over {\sqrt 2 }}$$
$$ = \left( {\sqrt {{3 \over 2}} - 1} \right)$$
Comments (0)
