IIT-JEE 2008 · Paper 2 · Mathematics Q3

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MathematicsSingle correct+3 / -1Question 3 of 66

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 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{\frac23}

\sqrt{\frac32}-1

1+\sqrt{\frac23}

\sqrt{\frac32}+1

Answer and worked solution

Official answer: B

Standard form. Completing squares:
$(x-\sqrt2)^2-2(y+\sqrt2)^2=4\implies\frac{(x-\sqrt2)^2}4-\frac{(y+\sqrt2)^2}2=1,$
so $a=2$ , $b=\sqrt2$ , $e=\sqrt{1+\frac24}=\sqrt{\frac32}$ .

Triangle. A (vertex) and C (nearest focus) lie on the axis, $AC=ae-a=2\left(\sqrt{\frac32}-1\right)$ . B is an end of the latus rectum through C, at distance $\frac{b^2}a=1$ from the axis. Right-angled at C:
$\text{Area}=\frac12\cdot2\left(\sqrt{\tfrac32}-1\right)\cdot1=\sqrt{\frac32}-1.$

(B)

Topics:Hyperbola

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IIT-JEE 2008 · Paper 2, Mathematics question 3