The area of the region bounded by the curve 9 | vedclass

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(D) The given equation is $9x^2 + 4y^2 - 36 = 0$.Rearranging the terms,we get $9x^2 + 4y^2 = 36$.Dividing both sides by $36$,we obtain $\frac{x^2}{4}

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$6$

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(D) The given equation is $9x^2 + 4y^2 - 36 = 0$.Rearranging the terms,we get $9x^2 + 4y^2 = 36$.Dividing both sides by $36$,we obtain $\frac{x^2}{4} + \frac{y^2}{9} = 1$.This is the standard equation of an ellipse $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,where $b^2 = 4$ and $a^2 = 9$.Thus,$b = 2$ and $a = 3$.The area of an ellipse is given by the formula $A = \pi ab$.Substituting the values,$A = \pi 	imes 2 	imes 3 = 6\pi$ square units.

The area of the region bounded by the curve $9x^2 + 4y^2 - 36 = 0$ is (in $\pi$)

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