Area of the ellipse frac{x^2}{a^2} + frac{y^2 | vedclass

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(A) The given equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the equation contains only even powers of $x$ and $y$,the ellip

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$\pi ab 	ext{ sq. unit}$

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(A) The given equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the equation contains only even powers of $x$ and $y$,the ellipse is symmetric about both the $x$-axis and the $y$-axis.Therefore,the total area of the ellipse is $4$ times the area of the region in the first quadrant (region $OBC$).Area $= 4 \int_{0}^{a} y \, dx = 4 \int_{0}^{a} \frac{b}{a} \sqrt{a^2 - x^2} \, dx$.Let $x = a \sin 	heta$,then $dx = a \cos 	heta \, d	heta$. When $x=0, 	heta=0$ and when $x=a, 	heta=\frac{\pi}{2}$.Area $= 4 \frac{b}{a} \int_{0}^{\pi/2} \sqrt{a^2 - a^2 \sin^2 	heta} \cdot a \cos 	heta \, d	heta = 4ab \int_{0}^{\pi/2} \cos^2 	heta \, d	heta$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:Area $= 4ab \int_{0}^{\pi/2} \frac{1 + \cos 2	heta}{2} \, d	heta = 2ab \left[ 	heta + \frac{\sin 2	heta}{2} ight]_{0}^{\pi/2}$.Area $= 2ab \left( (\frac{\pi}{2} + 0) - (0 + 0) ight) = \pi ab 	ext{ sq. unit}$.

Area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

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