Question

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 : 

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

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

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

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

Answer 1

Correct option is : (1) \(\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\) 

\(y^{2}=4 x\)

\(x^{2}+y^{2}=5\)

\(\therefore\) Area of shaded region as shown in the figure will be 

\(A_{1}=\int_{0}^{1} \sqrt{4 x} \ d x+\int_{1}^{\sqrt{5}} \sqrt{5-x^{2}} \ d x\)

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

\(=\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\)

\(\therefore\) Required Area \( =2 \mathrm{~A}_{1}\)

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

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

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

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