The area of the region, inside the ellipse x^ | vedclass

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(A) The given ellipse is $x^{2}+4y^{2}=4$, which can be written as $\frac{x^{2}}{4}+y^{2}=1$. Here, $a=2$ and $b=1$. The area of the ellipse is $A_{e}

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$2(\pi-1)$

Vedclass · Answer

(A) The given ellipse is $x^{2}+4y^{2}=4$, which can be written as $\frac{x^{2}}{4}+y^{2}=1$. Here, $a=2$ and $b=1$. The area of the ellipse is $A_{e} = \pi ab = \pi 	imes 2 	imes 1 = 2\pi$. The region bounded by the curves $y=|x|-1$ and $y=1-|x|$ is a square with vertices at $(1,0), (0,1), (-1,0),$ and $(0,-1)$. The side length of this square is $s = \sqrt{(1-0)^{2}+(0-1)^{2}} = \sqrt{1+1} = \sqrt{2}$. The area of this square is $A_{s} = s^{2} = (\sqrt{2})^{2} = 2$. The required area is the area inside the ellipse and outside the square, which is $A_{e} - A_{s} = 2\pi - 2 = 2(\pi-1)$.

The area of the region, inside the ellipse $x^{2}+4y^{2}=4$ and outside the region bounded by the curves $y=|x|-1$ and $y=1-|x|$, is:

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