Step 1: Analyze the equation of the ellipse

The given equation of the ellipse is $x^2 + 9y^2 = 9$. To convert this to the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we divide the entire equation by 9: \frac{x^2}{9} + \frac{9y^2}{9} = \frac{9}{9} $$$$ \frac{x^2}{3^2} + \frac{y^2}{1^2} = 1Comparing this with the standard form, we find the semi-major axis $a = 3$ and the semi-minor axis $b = 1$. The major axis is along the x-axis, and the minor axis is along the y-axis.

Step 2: Identify the coordinates of points A and B

• Point $A$ is an extremity of the major axis. The extremities of the major axis are $(\pm a, 0)$. We can choose $A = (3, 0)$.
• Point $B$ is an extremity of the minor axis. The extremities of the minor axis are $(0, \pm b)$. We can choose $B = (0, 1)$.

(Note: The choice of which extremities to use for $A$ and $B$, e.g., $(-3,0)$ or $(0,-1)$, will result in a symmetric problem and yield the same area for the triangle.)

Step 3: Find the equation of the line passing through A and B

The line passes through $A(3, 0)$ and $B(0, 1)$. Using the intercept form of a line, $\frac{x}{x_{int}} + \frac{y}{y_{int}} = 1$, we get: \frac{x}{3} + \frac{y}{1} = 1 $$$$ x + 3y = 3This is the equation of the line AB.

Step 4: Determine the equation of the auxiliary circle

The auxiliary circle of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by the equation $x^2 + y^2 = a^2$. For the given ellipse, $a = 3$, so the equation of the auxiliary circle is: x^2 + y^2 = 3^2 $$$$ x^2 + y^2 = 9

Step 5: Find the coordinates of the intersection point M

The point $M$ is the intersection of the line AB and the auxiliary circle. We need to solve the system of two equations:

1. $x + 3y = 3 \implies x = 3 - 3y$
2. $x^2 + y^2 = 9$

Substitute the expression for $x$ from equation (1) into equation (2): (3 - 3y)^2 + y^2 = 9 $$$$ (9 - 18y + 9y^2) + y^2 = 9 $$$$ 10y^2 - 18y = 0 $$$$ 2y(5y - 9) = 0This gives two possible values for $y$: $y = 0$ or $y = \frac{9}{5}$.

• If $y = 0$, then $x = 3 - 3(0) = 3$. This corresponds to the point $(3, 0)$, which is point $A$.
• If $y = \frac{9}{5}$, then $x = 3 - 3(\frac{9}{5}) = 3 - \frac{27}{5} = \frac{15 - 27}{5} = -\frac{12}{5}$.

So, the coordinates of the point $M$ are $(-\frac{12}{5}, \frac{9}{5})$.

Step 6: Calculate the area of the triangle AOM

The vertices of the triangle are $A(3, 0)$, $O(0, 0)$, and $M(-\frac{12}{5}, \frac{9}{5})$.

We can calculate the area of a triangle with one vertex at the origin using the formula: Area = $\frac{1}{2} |x_1 y_2 - x_2 y_1|$. Let $(x_1, y_1) = A(3, 0)$ and $(x_2, y_2) = M(-\frac{12}{5}, \frac{9}{5})$. \text{Area}(\triangle AOM) = \frac{1}{2} \left| (3)\left(\frac{9}{5}\right) - (0)\left(-\frac{12}{5}\right) \right| $$$$ \text{Area}(\triangle AOM) = \frac{1}{2} \left| \frac{27}{5} - 0 \right| $$$$ \text{Area}(\triangle AOM) = \frac{1}{2} \times \frac{27}{5} = \frac{27}{10} Alternatively, we can consider $OA$ as the base of the triangle. The length of the base $OA$ is 3 units (along the x-axis). The height of the triangle is the perpendicular distance from $M$ to the x-axis, which is the absolute value of the y-coordinate of $M$. Base = $OA = 3$ Height = $|y_M| = \frac{9}{5}$ $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10}$

Step 7: Conclusion

The area of the triangle with vertices at A, M, and the origin O is $\frac{27}{10}$. This matches option D.