Area (in sq. units) of region outside frac{|mathrm{x}|}{2}+frac{|mathrm{y}|}{3}=1 and inside ellipse

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10-2. Parabola, Ellipse, Hyperbola

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Area (in sq. units) of the region outside $\frac{|\mathrm{x}|}{2}+\frac{|\mathrm{y}|}{3}=1$ and inside the ellipse $\frac{\mathrm{x}^{2}}{4}+\frac{\mathrm{y}^{2}}{9}=1$ is

A

$3(4-\pi)$

B

$6(\pi-2)$

C

$3(\pi-2)$

D

$6(4-\pi)$

(JEE MAIN-2020)

Solution

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

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

Area of Ellipse $=\pi a b=6 \pi$

Required area,

$=\pi \times 2 \times 3-($ Area of quadrilateral $)$

$=6 \pi-\frac{1}{2} 6 \times 4$

$=6 \pi-12$

$=6(\pi-2)$

Standard 11

Mathematics

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