We have,

$x^2+4y^2=5$

$\Rightarrow \frac{x^2}{5}+\frac{y^2}{\left({\displaystyle \frac{5}{4}}\right)}=1$

Tangent at $P\left(1,1\right)$ to the ellipse is

$x+4y=5$
Required Area

$={\int }_1^{\sqrt{5}}\left(\frac{5-x}{4}-\frac{\sqrt{5-x^2}}{2}\right)dx$

$={\left[\frac{5x}{4}-\frac{x^2}{8}-\frac{x}{4}\sqrt{5-x^2}-\frac{5}{4}{\sin }^{-1}\frac{x}{\sqrt{5}}\right]}_1^{\sqrt{5}}$

$=\left[\frac{5\sqrt{5}}{4}-\frac{5}{8}-\frac{5}{4}·\left(\frac{\mathrm{\pi }}{2}\right)-\frac{5}{4}+\frac{1}{8}+\frac{1}{2}+\frac{5}{4}{\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]$

$=\left[\frac{5\sqrt{5}}{4}-\frac{5}{4}-\frac{5}{4}·\left(\frac{\mathrm{\pi }}{2}\right)+\frac{5}{4}\left(\frac{\mathrm{\pi }}{2}-{\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)\right]$

$=\left[\frac{5\sqrt{5}}{4}-\frac{5}{4}-\frac{5\mathrm{\pi }}{8}+\frac{5\mathrm{\pi }}{8}-\frac{5}{4}{\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)\right]$

$=\frac{5}{4}\sqrt{5}-\frac{5}{4}-\frac{5}{4}{\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)$

So,

$\alpha =\frac{5}{4}, \beta =-\frac{5}{4} \& \gamma =-\frac{5}{4}$

$\left|\alpha +\beta +\gamma \right|=1.25$