We have,
x2+4y2=5
⇒5x2+(45)y2=1

Tangent at P(1,1) to the ellipse is
x+4y=5
Required Area
=∫15(45−x−25−x2)dx
=[45x−8x2−4x5−x2−45sin−15x]15
=[455−85−45⋅(2π)−45+81+21+45sin−1(51)]
=[455−45−45⋅(2π)+45(2π−cos−1(51))]
=[455−45−85π+85π−45cos−1(51)]
=455−45−45cos−1(51)
So,
\alpha =\frac{5}{4},\beta =-\frac{5}{4}&\gamma =-\frac{5}{4}
∣α+β+γ∣=1.25