The parabola y 2 = 4 x divides the area of the circle x 2 + y 2 = 5 in two parts. The area of the smaller part is equal to : [2024]

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Solution

Q.
The parabola $y^{2} = 4 x$ divides the area of the circle $x^{2} + y^{2} = 5$ in two parts. The area of the smaller part is equal to : [2024]

1 $\frac{1}{3} + 5 \sin^{- 1} (\frac{2}{\sqrt{5}})$

2 $\frac{2}{3} + \sqrt{5} \sin^{- 1} (\frac{2}{\sqrt{5}})$

3 $\frac{2}{3} + 5 \sin^{- 1} (\frac{2}{\sqrt{5}})$

4 $\frac{1}{3} + \sqrt{5} \sin^{- 1} (\frac{2}{\sqrt{5}})$

Ans.
(3)

The points of intersection of $y^{2} = 4 x$ and $x^{2} + y^{2} = 5$ are (1, 2) and (1, -2).

$∴$ $Required Area = 2 {Area of OACO + Area of CABC}$

$= 2 [\int_{0}^{1} 2 \sqrt{x} d x + \int_{1}^{\sqrt{5}} \sqrt{5 - x^{2}} d x]$

$= 2 {[| \frac{4}{3} x^{\frac{3}{2}} |_{0}^{1} + (\frac{1}{2} x \sqrt{5 - x^{2}} + \frac{5}{2} \sin^{- 1} \frac{x}{\sqrt{5}})]}_{1}^{\sqrt{5}}$

$= 2 [(\frac{4}{3} - 0) + (0 + \frac{5 π}{4}) - (1 + \frac{5}{2} \sin^{- 1} \frac{1}{\sqrt{5}})]$

$= 2 [\frac{1}{3} + \frac{5 π}{4} - \frac{5}{2} \sin^{- 1} \frac{1}{\sqrt{5}}] = \frac{2}{3} + 5 [\frac{π}{2} - \sin^{- 1} \frac{1}{\sqrt{5}}]$

$= \frac{2}{3} + 5 \cos^{- 1} (\frac{1}{\sqrt{5}})$ $[∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}]$

$= \frac{2}{3} + 5 \sin^{- 1} (\frac{2}{\sqrt{5}})$

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