In the figure,AOBA is the part of the ellipse | vedclass

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(A) The given equation of the ellipse is $9x^{2} + y^{2} = 36$,which can be written as $\frac{x^{2}}{4} + \frac{y^{2}}{36} = 1$ or $\frac{x^{2}}{2^{2}

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$3\pi - 6$

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(A) The given equation of the ellipse is $9x^{2} + y^{2} = 36$,which can be written as $\frac{x^{2}}{4} + \frac{y^{2}}{36} = 1$ or $\frac{x^{2}}{2^{2}} + \frac{y^{2}}{6^{2}} = 1$.From the equation,$y = \sqrt{36 - 9x^{2}} = 3\sqrt{4 - x^{2}}$.The equation of the chord $AB$ passing through $A(2, 0)$ and $B(0, 6)$ is given by $\frac{x}{2} + \frac{y}{6} = 1$,which simplifies to $y = 6 - 3x$.The area of the shaded region is the area under the ellipse minus the area under the line $AB$ from $x = 0$ to $x = 2$.Area $= \int_{0}^{2} (y_{	ext{ellipse}} - y_{	ext{line}}) dx = \int_{0}^{2} (3\sqrt{4 - x^{2}} - (6 - 3x)) dx$.$= 3 \int_{0}^{2} \sqrt{2^{2} - x^{2}} dx - \int_{0}^{2} (6 - 3x) dx$.Using the formula $\int \sqrt{a^{2} - x^{2}} dx = \frac{x}{2}\sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2}\sin^{-1}(\frac{x}{a})$:$= 3 \left[ \frac{x}{2}\sqrt{4 - x^{2}} + \frac{4}{2}\sin^{-1}(\frac{x}{2}) ight]_{0}^{2} - \left[ 6x - \frac{3x^{2}}{2} ight]_{0}^{2}$.$= 3 \left[ (0 + 2\sin^{-1}(1)) - (0 + 0) ight] - \left[ (12 - 6) - (0 - 0) ight]$.$= 3 \left[ 2 	imes \frac{\pi}{2} ight] - 6 = 3\pi - 6$ square units.

In the figure,$AOBA$ is the part of the ellipse $9x^{2} + y^{2} = 36$ in the first quadrant such that $OA = 2$ and $OB = 6$. Find the area between the arc $AB$ and the chord $AB$.

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