Question

The area of the region bounded by the inequalities $4<x^2+y^2<16$ and $3x^2-y^2>0$ is $d.$ Then the value of $\frac{d}{\pi }$ is

Answer 1

The points satisfying inequalities
$4<x^2+y^2<16$ lie in the region outside the circle $x^2+y^2=4$ and inside the circle $x^2+y^2=16$
Now $3x^2-y^2>0$
$\therefore (\sqrt{3}x-y)(\sqrt{3}x+y)>0$
So the region of the points satisfying both the inequalities is

Required area $=4\times \frac{\frac{\pi }{3}}{2\pi }[\pi (4{)}^2-\pi (2{)}^2]=8\pi =d\Rightarrow \frac{d}{\pi }=8$