Question

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

Answer 1

$E:x^2+4y^2=5$
Equation of tangent at (1,1) is $x+4y=5$
Required area $=$ Area of trapezium $SPQA–$ Area under the segment of ellipse


$={\int }_1^{\sqrt{5}}\{\left(\frac{5-x}{4}\right)-\frac{1}{2}\sqrt{5-x^2}\}dx=\frac{5\sqrt{5}-7}{4}-\frac{1}{4}{[x\sqrt{5-x^2}+5{\sin }^{-1}\left(\frac{x}{\sqrt{5}}\right)]}_1^{\sqrt{5}}=\frac{5\sqrt{5}-7}{4}-\frac{5{\text{cos}}^{-1}\left(\frac{1}{\sqrt{5}}\right)-2}{4}=\frac{5\sqrt{5}}{4}-\frac{5}{4}-\frac{5}{4}{\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Clearly, $\alpha =\frac{5}{4},\beta =-\frac{5}{4}$ and $\gamma =-\frac{5}{4}$
$\Rightarrow |\alpha +\beta +\gamma |=\frac{5}{4}$