Let T be the tangent to the ellipse E:x^2+4y^2=5 at the point P(1,1).

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 $T$, the ellipse $E$, and the lines $x=1$ and $x=\sqrt5$ is $\alpha\sqrt5+\beta+\gamma\cos^{-1}\dfrac1{\sqrt5}$, then $|\alpha+\beta+\gamma|$ is equal to ......

Answer: 1.25

Accepted Answer

Tangent at $(1,1)$: $x+4y=5$.

$$\text{Area}=\int_1^{\sqrt5}\left[\frac{5-x}4-\frac12\sqrt{5-x^2}\right]dx.$$

Evaluating (using $\int\sqrt{5-x^2}\,dx=\frac x2\sqrt{5-x^2}+\frac52\sin^{-1}\frac x{\sqrt5}$) and simplifying with $\sin^{-1}\frac1{\sqrt5}=\frac\pi2-\cos^{-1}\frac1{\sqrt5}$:

$$\text{Area}=\frac54\sqrt5-\frac54-\frac54\cos^{-1}\frac1{\sqrt5}.$$

$$\alpha=\frac54,\ \beta=-\frac54,\ \gamma=-\frac54,\qquad|\alpha+\beta+\gamma|=\frac54=1.25.$$

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