JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Area Under Curves

Completing the square, $16(x+2{)}^2-(y-2{)}^2=16$, i.e. $\frac{(x+2{)}^2}{1}-\frac{(y-2{)}^2}{16}=1$, a hyperbola centred at $(-2,2)$ whose transverse axis is the line $y=2$ and conjugate axis the line $x=-2$. The parabola is $y=x^2-4$, meeting $y=2$ at $x=\pm \sqrt{6}$. The region lies to the right of $x=-2$, below $y=2$ and above the parabola, so its area is ${\int }_{-2}^{\sqrt{6}}(2-(x^2-4))dx={\int }_{-2}^{\sqrt{6}}(6-x^2)dx=[6x-\frac{x^3}{3}{]}_{-2}^{\sqrt{6}}=4\sqrt{6}+\frac{28}{3}$.