Area (in sq. units) of the region outside x 2 + y 3 = 1 and inside the ellipse x 2 4 + y 2 9 = 1 is

Area (in sq. units) of the region outside

\frac{|x|}{2} + \frac{|y|}{3} = 1

and inside the ellipse

\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1

is

$6 (π - 2)$
$3 (π - 2)$
$3 (4 - π)$
$6 (4 - π)$

Solution

Area of ellipse $= π a b = 6 π$
Area enclosed by $\frac{| x |}{2} + \frac{| y |}{3} = 1$ is
$= 4 \times \frac{1}{2} (2) (3) = 12$
So, required area $= 6 π - 12$ sq. units

Asked in: JEE Main 2020 (02 Sep Shift 1)

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