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…

We have,

$${x}^{2}+4{y}^{2}=5$$x2+4y2=5

$$\Rightarrow \frac{{x}^{2}}{5}+\frac{{y}^{2}}{(\frac{5}{4})}=1$$⇒5x2​+(45​)y2​=1

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

$$x+4y=5$$x+4y=5

Required Area

$$={\int }_{1}^{\sqrt{5}}(\frac{5-x}{4}-\frac{\sqrt{5-{x}^{2}}}{2})dx$$=∫15​​(45−x​−25−x2​​)dx

$$={[\frac{5x}{4}-\frac{{x}^{2}}{8}-\frac{x}{4}\sqrt{5-{x}^{2}}-\frac{5}{4}{\mathrm{sin}}^{-1}\frac{x}{\sqrt{5}}]}_{1}^{\sqrt{5}}$$=[45x​−8x2​−4x​5−x2​−45​sin−15​x​]15​​

$$=[\frac{5\sqrt{5}}{4}-\frac{5}{8}-\frac{5}{4}\cdot (\frac{\pi }{2})-\frac{5}{4}+\frac{1}{8}+\frac{1}{2}+\frac{5}{4}{\mathrm{sin}}^{-1}(\frac{1}{\sqrt{5}})]$$=[455​​−85​−45​⋅(2π​)−45​+81​+21​+45​sin−1(5​1​)]

$$=[\frac{5\sqrt{5}}{4}-\frac{5}{4}-\frac{5}{4}\cdot (\frac{\pi }{2})+\frac{5}{4}(\frac{\pi }{2}-{\mathrm{cos}}^{-1}(\frac{1}{\sqrt{5}}))]$$=[455​​−45​−45​⋅(2π​)+45​(2π​−cos−1(5​1​))]

$$=[\frac{5\sqrt{5}}{4}-\frac{5}{4}-\frac{5\pi }{8}+\frac{5\pi }{8}-\frac{5}{4}{\mathrm{cos}}^{-1}(\frac{1}{\sqrt{5}})]$$=[455​​−45​−85π​+85π​−45​cos−1(5​1​)]

$$=\frac{5}{4}\sqrt{5}-\frac{5}{4}-\frac{5}{4}{\mathrm{cos}}^{-1}(\frac{1}{\sqrt{5}})$$=45​5​−45​−45​cos−1(5​1​)

So,

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

$$|\alpha +\beta +\gamma |=1.25$$∣α+β+γ∣=1.25