The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse. x2 + 9y2 = 9, meets the auxillary circle at the point M, then the area of the triangle with vertices at A, M and the origin O is

Equation of the ellipse is x29+y21=1and of theauxiliary circle is x2 + y2 = 9→(1)Equation of AB isx3+y1=1 ⇒x3=1-y ⇒x=3(1-y)y-coordinate of M is given bysub x in (1)[3(1−y)]2+y2=9⇒10y2−18y=0⇒y=0 or y=95Area of ΔOAM=12(OA)(MN)=12×3×95=2710

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The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse. x²+ 9y² = 9, meets the auxillary 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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Detailed Solution

Equation of the ellipse is $\frac{x^{2}}{9} + \frac{y^{2}}{1} = 1$ and of the
auxiliary circle is x²+ y² = 9 $\to (1)$
Equation of AB is $\frac{x}{3} + \frac{y}{1} = 1$
$\Rightarrow \frac{x}{3} = 1 - y \Rightarrow x = 3 (1 - y)$

y-coordinate of M is given by

sub x in (1)

$[3 (1 - y)]^{2} + y^{2} = 9$
$\begin{array}{l} \Rightarrow 10 y^{2} - 18 y = 0 \Rightarrow y = 0 or y = \frac{9}{5} \\ A r e a o f ΔOAM = \frac{1}{2} (OA) (MN) = \frac{1}{2} \times 3 \times \frac{9}{5} = \frac{27}{10} \end{array}$