Area lying in the first quadrant and bounded | vedclass

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(D) The given equation of the ellipse is $4x^2 + 9y^2 = 144$.Dividing both sides by $144$,we get $\frac{4x^2}{144} + \frac{9y^2}{144} = 1$,which simpl

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$6$

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(D) The given equation of the ellipse is $4x^2 + 9y^2 = 144$.Dividing both sides by $144$,we get $\frac{4x^2}{144} + \frac{9y^2}{144} = 1$,which simplifies to $\frac{x^2}{36} + \frac{y^2}{16} = 1$.This is the standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 36$ (so $a = 6$) and $b^2 = 16$ (so $b = 4$).The total area of an ellipse is given by the formula $A = \pi ab$.Substituting the values,the total area is $A = \pi 	imes 6 	imes 4 = 24\pi$.Since the ellipse is symmetric about both axes,the area in the first quadrant is one-fourth of the total area.Therefore,the required area = $\frac{1}{4} 	imes 24\pi = 6\pi$.

Area lying in the first quadrant and bounded by the ellipse $4x^2 + 9y^2 = 144$ is . . . . . . . (in $\pi$)

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