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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
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The correct answer is:
$\frac{27}{10}$
$\frac{27}{10}$
Equation of auxiliary circle is
$$
x^2+y^2=9
$$
Equation of $A M$ is $\frac{x}{3}+\frac{y}{1}=1$

On solving Eqs. (i) and (ii), we get Coordinates of $M=\left(-\frac{12}{5}, \frac{9}{5}\right)$ Now, area of $\triangle A O M=\frac{1}{2} \times O A \times M N$ $=\frac{27}{10}$ sq units
$$
x^2+y^2=9
$$
Equation of $A M$ is $\frac{x}{3}+\frac{y}{1}=1$

On solving Eqs. (i) and (ii), we get Coordinates of $M=\left(-\frac{12}{5}, \frac{9}{5}\right)$ Now, area of $\triangle A O M=\frac{1}{2} \times O A \times M N$ $=\frac{27}{10}$ sq units
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