Question:
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 :
Correct Option: , 2
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 $)$
$=6 \pi-\frac{1}{2} 6 \times 4$
$=6 \pi-12$
$=6(\pi-2)$
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