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
Answer & explanation
Correct answer: option 3
Let A be the required area. Then,
A = Area of the ellipse - Area of rhombus ABA' B'
$⇒A=π(2) (3)-\frac{1}{2}(AA' × BB')$
$⇒A=6π-\frac{1}{2}(4×6)=6π-12=6 (π-2)$