JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 12)
The line passing through the extremity $$A$$ of the major axis and extremity $$B$$ of the minor axis of the ellipse $${x^2} + 9{y^2} = 9$$ meets its auxiliary circle at the point $$M$$. Then the area of the triangle with vertices at $$A$$, $$M$$ and the origin $$O$$ is
$${{31} \over {10}}$$
$${{29} \over {10}}$$
$${{21} \over {10}}$$
$${{27} \over {10}}$$
Explanation
Equation of line AM is $$x + 3y - 3 = 0$$.
Perpendicular distance of line from origin is $$3/\sqrt {10} $$.
Length of AM is
$$2\sqrt {9 - {9 \over {10}}} = 2 \times {9 \over {\sqrt {10} }}$$
The required area of the rectangle is
$${1 \over 2} \times 2 \times {9 \over {\sqrt {10} }} \times {3 \over {\sqrt {10} }} = {{27} \over {10}}$$ sq. unit
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