The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^{2} + 9y^{2} = 9$ meets its auxiliary circle at the point $M$. Then the area of the triangle with vertices at $A$, $M$ and the origin $O$ is

$\text{Given: ellipse } x^2 + 9y^2 = 9$

$\text{line through } A \text{ (major axis end) and } B \text{ (minor axis end)}$

To find: area of triangle AMO (with M on the auxiliary circle)

$\dfrac{x^2}{9} + \dfrac{y^2}{1} = 1$

$a = 3, \quad b = 1$

$A(3, 0), \quad B(0, 1)$

$\text{Auxiliary circle: } x^2 + y^2 = 9$

$\dfrac{x}{3} + \dfrac{y}{1} = 1 \ \Rightarrow \ x + 3y = 3$

$x = 3 - 3y$

$(3-3y)^2 + y^2 = 9 \ \Rightarrow \ 10y^2 - 18y = 0$

$2y(5y - 9) = 0 \ \Rightarrow \ y = 0 \ \text{or} \ y = \dfrac{9}{5}$

$M\left(-\dfrac{12}{5}, \dfrac{9}{5}\right)$

[triangle AMO] = 1/2| x_A y_M - x_M y_A |

$= \dfrac{1}{2}\left| 3 \cdot \dfrac{9}{5} - \left(-\dfrac{12}{5}\right)\cdot 0 \right| = \dfrac{1}{2}\cdot\dfrac{27}{5} = \dfrac{27}{10}$

$\boxed{\text{Area} = \dfrac{27}{10} \ \text{— option (4)}}$