JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Ellipse

The ellipse is $\frac{x^2}{9}+\frac{y^2}{1}=1$, so $a=3$, $b=1$. The extremity of the major axis is $A(3,0)$ and the extremity of the minor axis is $B(0,1)$.

The equation of line $AB$ is: $\frac{x}{3}+\frac{y}{1}=1$, or $x+3y-3=0$.

The auxiliary circle is $x^2+y^2=9$ (radius $a=3$).

The perpendicular distance from origin $O$ to line $AB$ is:


The length of $AM$ can be found using the chord length formula. The distance from center to chord is $\frac{3}{\sqrt{10}}$, so:


Area of triangle $OAM=\frac{1}{2}\times AM\times d=\frac{1}{2}\times \frac{18}{\sqrt{10}}\times \frac{3}{\sqrt{10}}=\frac{27}{10}$