Let $$T$$ be the tangent to the ellipse $$E : x^2 + 4y^2 = 5$$ at the point $$P(1, 1)$$. If the area of the region bounded by the tangent $$T$$, ellipse $$E$$, lines $$x = 1$$ and $$x = \sqrt{5}$$ is $$\alpha\sqrt{5} + \beta + \gamma\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)$$, then $$|\alpha + \beta + \gamma|$$ is equal to ___.

The tangent to ellipse $$x^2 + 4y^2 = 5$$ at $$P(1,1)$$ is $$x \cdot 1 + 4y \cdot 1 = 5$$, i.e., $$x + 4y = 5$$, giving $$y = \frac{5-x}{4}$$.

The upper half of the ellipse gives $$y = \frac{1}{2}\sqrt{5 - x^2}$$. The region is bounded between $$x = 1$$ and $$x = \sqrt{5}$$, with the tangent line above the ellipse in this interval (both meet at $$x=1$$ and the ellipse hits $$y=0$$ at $$x=\sqrt{5}$$ while the tangent is still positive there).

$$\text{Area} = \int_1^{\sqrt{5}} \left[\frac{5-x}{4} - \frac{1}{2}\sqrt{5-x^2}\right]dx$$

For the first integral: $$\int_1^{\sqrt{5}}\frac{5-x}{4}\,dx = \frac{1}{4}\left[5x - \frac{x^2}{2}\right]_1^{\sqrt{5}} = \frac{1}{4}\left[\left(5\sqrt{5} - \frac{5}{2}\right) - \left(5 - \frac{1}{2}\right)\right] = \frac{5\sqrt{5} - 7}{4}$$

For the second integral, using $$\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C$$ with $$a^2=5$$:

$$\int_1^{\sqrt{5}}\frac{1}{2}\sqrt{5-x^2}\,dx = \frac{1}{2}\left[\frac{x}{2}\sqrt{5-x^2}+\frac{5}{2}\arcsin\frac{x}{\sqrt{5}}\right]_1^{\sqrt{5}} = \frac{1}{2}\left[\frac{5\pi}{4} - 1 - \frac{5}{2}\arcsin\frac{1}{\sqrt{5}}\right]$$

Using $$\arcsin\frac{1}{\sqrt{5}} = \frac{\pi}{2} - \cos^{-1}\frac{1}{\sqrt{5}}$$, the second integral becomes $$\frac{5\pi}{8} - \frac{1}{2} - \frac{5}{4}\left(\frac{\pi}{2} - \cos^{-1}\frac{1}{\sqrt{5}}\right) = -\frac{1}{2} + \frac{5}{4}\cos^{-1}\frac{1}{\sqrt{5}}$$.

$$\text{Area} = \frac{5\sqrt{5}-7}{4} - \left(-\frac{1}{2} + \frac{5}{4}\cos^{-1}\frac{1}{\sqrt{5}}\right) = \frac{5}{4}\sqrt{5} - \frac{7}{4} + \frac{2}{4} - \frac{5}{4}\cos^{-1}\frac{1}{\sqrt{5}}$$

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

Matching with $$\alpha\sqrt{5} + \beta + \gamma\cos^{-1}\!\left(\frac{1}{\sqrt{5}}\right)$$, we get $$\alpha = \frac{5}{4}$$, $$\beta = -\frac{5}{4}$$, $$\gamma = -\frac{5}{4}$$.

$$|\alpha + \beta + \gamma| = \left|\frac{5}{4} - \frac{5}{4} - \frac{5}{4}\right| = \frac{5}{4} = 1.25$$