JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Area Under Curves

The tangent to $x^2+4y^2=5$ at $P(1,1)$ is $x+4y=5$, while the upper half of the ellipse is $y=\frac{1}{2}\sqrt{5-x^2}$; the two meet at $x=1$ and the ellipse reaches the axis at $x=\sqrt{5}$. The region between them is The straight part gives $\frac{1}{4}{\left[5x-\frac{x^2}{2}\right]}_1^{\sqrt{5}}=\frac{5\sqrt{5}}{4}-\frac{7}{4}$, and with $\int \sqrt{5-x^2}dx=\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}{\sin }^{-1}\frac{x}{\sqrt{5}}$ the elliptic part gives $\frac{5\pi }{8}-\frac{1}{2}-\frac{5}{4}{\sin }^{-1}\frac{1}{\sqrt{5}}$. Subtracting and using ${\sin }^{-1}\frac{1}{\sqrt{5}}=\frac{\pi }{2}-{\cos }^{-1}\frac{1}{\sqrt{5}}$, the multiples of $\pi$ cancel and Hence $\alpha =\frac{5}{4}$, $\beta =-\frac{5}{4}$, $\gamma =-\frac{5}{4}$, so $|\alpha +\beta +\gamma |=\frac{5}{4}$ and $4|\alpha +\beta +\gamma |=5$.