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

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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 length of the semi-major axis is $a=3$ and the length of 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 equation of the auxiliary circle of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2$,so $x^2+y^2=9$.The line $AB$ intersects the circle $x^2+y^2=9$ at point $M$. Substituting $x=3-3y$ into the circle equation:$(3-3y)^2+y^2=9$$9-18y+9y^2+y^2=9$$10y^2-18y=0$$2y(5y-9)=0$.Since $y=0$ corresponds to point $A(3,0)$,the point $M$ has $y=\frac{9}{5}$.Then $x=3-3(\frac{9}{5}) = 3-\frac{27}{5} = -\frac{12}{5}$.So,$M = \left(-\frac{12}{5}, \frac{9}{5}\right)$.The area of triangle $AMO$ with vertices $A(3,0)$,$M(-\frac{12}{5}, \frac{9}{5})$,and $O(0,0)$ is given by:Area $= \frac{1}{2} |x_A(y_M-y_O) + x_M(y_O-y_A) + x_O(y_A-y_M)|$Area $= \frac{1}{2} |3(\frac{9}{5}-0) + (-\frac{12}{5})(0-0) + 0(0-\frac{9}{5})|$Area $= \frac{1}{2} |\frac{27}{5}| = \frac{27}{10}$.

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

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