Question:

<p><strong>Step 1: Rewrite the ellipse in standard form.</strong></p><p>Given: $x^2 + 9y^2 = 9$</p><p>Dividing by 9: $\frac{x^2}{9} + \frac{y^2}{1} = 1$</p><p>So $a^2 = 9$, $b^2 = 1$, giving $a = 3$, $b = 1$</p><p>Major axis is along the x-axis, minor axis along y-axis.</p><p><strong>Step 2: Identify points A and B.</strong></p><p>Extremity A of major axis: $A = (3, 0)$</p><p>Extremity B of minor axis: $B = (0, 1)$ (taking the positive extremity)</p><p><strong>Step 3: Find the equation of line AB.</strong></p><p>Line through $A(3, 0)$ and $B(0, 1)$:</p><p>$\frac{x}{3} + \frac{y}{1} = 1$</p><p>Or: $x + 3y = 3$</p><p><strong>Step 4: Find the auxiliary circle equation.</strong></p><p>The auxiliary circle of the ellipse has radius equal to the semi-major axis.</p><p>Equation: $x^2 + y^2 = 9$</p><p><strong>Step 5: Find intersection point M.</strong></p><p>Substitute $x = 3 - 3y$ into $x^2 + y^2 = 9$:</p><p>$(3 - 3y)^2 + y^2 = 9$</p><p>$9 - 18y + 9y^2 + y^2 = 9$</p><p>$10y^2 - 18y = 0$</p><p>$y(10y - 18) = 0$</p><p>So $y = 0$ or $y = \frac{18}{10} = \frac{9}{5}$</p><p>When $y = 0$: $x = 3$ (this gives point A)</p><p>When $y = \frac{9}{5}$: $x = 3 - 3 \cdot \frac{9}{5} = 3 - \frac{27}{5} = \frac{15 - 27}{5} = -\frac{12}{5}$</p><p>So $M = \left(-\frac{12}{5}, \frac{9}{5}\right)$</p><p><strong>Step 6: Calculate the area of triangle AOM.</strong></p><p>With vertices $O(0,0)$, $A(3, 0)$, and $M\left(-\frac{12}{5}, \frac{9}{5}\right)$:</p><p>Area $= \frac{1}{2}|x_A(y_M - y_O) + x_M(y_O - y_A) + x_O(y_A - y_M)|$</p><p>$= \frac{1}{2}\left|3 \cdot \frac{9}{5} + \left(-\frac{12}{5}\right) \cdot 0 + 0\right|$</p><p>$= \frac{1}{2} \cdot \frac{27}{5} = \frac{27}{10}$</p><p><strong>∴ Answer: D</strong></p>