Area enclosed by the ellipse 9x^2 + 4y^2 = 36 | vedclass

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(C) The given equation of the ellipse is $9x^2 + 4y^2 = 36$. Dividing both sides by $36$, we get: $\frac{9x^2}{36} + \frac{4y^2}{36} = 1$ $\frac{x^2}{

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$6$

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(C) The given equation of the ellipse is $9x^2 + 4y^2 = 36$. Dividing both sides by $36$, we get: $\frac{9x^2}{36} + \frac{4y^2}{36} = 1$ $\frac{x^2}{4} + \frac{y^2}{9} = 1$ This is of the form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a^2 = 9$ and $b^2 = 4$. So, $a = 3$ and $b = 2$. The area of an ellipse is given by the formula $A = \pi ab$. Substituting the values, we get: $A = \pi 	imes 3 	imes 2 = 6\pi$ sq. units.

Area enclosed by the ellipse $9x^2 + 4y^2 = 36$ is . . . . . . sq. units. (in $\pi$)

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