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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
$\frac{31}{10}$
$\frac{29}{10}$
$\frac{21}{10}$
$\frac{27}{10}$
Solution
The equation of ellipse is $\frac{x^2}{9}+y^2=1$
The length of semi-major axis is $a =3$ and the length of semi-minor axis is $b =1$
The coordinates of point $A$ is $(3,0)$ and the coordinates of point $B$ is $(0,1)$
The equation of line passing through $A , B$ is $x +3 y =3$
The equation of an auxiliary circle of an ellipse is $x^2+y^2=9$
The line $A B$ cuts $x^2+y^2=9$ at point $M$
By solving the above equations
The coordinates of point M are $\left(-\frac{12}{5}, \frac{9}{5}\right)$
The area of triangle AMO is $\frac{1}{2}\left|o\left(0-\frac{9}{5}\right)+3\left(\frac{9}{5}-0\right)-\frac{12}{5}(0-0)\right|=\frac{27}{10}$
