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(D) The equation of the ellipse is $\frac{x^2}{9} + y^2 = 1$. The semi-major axis is $a = 3$ and the semi-minor axis is $b = 1$. The coordinates of po

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$\frac{27}{10}$

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(D) The equation of the ellipse is $\frac{x^2}{9} + y^2 = 1$. The semi-major axis is $a = 3$ and the semi-minor axis is $b = 1$. The coordinates of point $A$ are $(3, 0)$ and the coordinates of point $B$ are $(0, 1)$. The equation of the line passing through $A$ and $B$ is $\frac{x}{3} + \frac{y}{1} = 1$,which simplifies to $x + 3y = 3$. The auxiliary circle of the ellipse is $x^2 + y^2 = a^2 = 9$. To find point $M$,we solve the system $x + 3y = 3$ and $x^2 + y^2 = 9$. From the first equation,$x = 3 - 3y$. Substituting into the second: $(3 - 3y)^2 + y^2 = 9$. $9 - 18y + 9y^2 + y^2 = 9 \implies 10y^2 - 18y = 0$. Thus,$y = 0$ (which gives $x = 3$,point $A$) or $y = \frac{18}{10} = \frac{9}{5}$. If $y = \frac{9}{5}$,then $x = 3 - 3(\frac{9}{5}) = 3 - \frac{27}{5} = -\frac{12}{5}$. So,$M = (-\frac{12}{5}, \frac{9}{5})$. The area of $	riangle AMO$ with vertices $O(0, 0)$,$A(3, 0)$,and $M(-\frac{12}{5}, \frac{9}{5})$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Area $= \frac{1}{2} |0(0 - \frac{9}{5}) + 3(\frac{9}{5} - 0) + (-\frac{12}{5})(0 - 0)| = \frac{1}{2} |3 	imes \frac{9}{5}| = \frac{27}{10}$.

$A$ line passing through the endpoint of the major axis $A$ and the endpoint of the minor axis $B$ of an ellipse $\frac{x^2}{9} + y^2 = 1$ touches its auxiliary circle at point $M$. Find the area of the triangle with vertices $A, M,$ and the origin $O$.

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