ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 8. Application and integration

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. A  \(\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
  2. B  \(\frac{1}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
  3. C \(\frac{1}{3}+\sqrt{5} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
  4. D \(\frac{2}{3}+\sqrt{5} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\( y^2=4 x \) \( x^2+y^2=5\) \(\therefore\) Area of shaded region as shown in the figure will be \( \mathrm{A}_1=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{dx}+\int_1^{\sqrt{5}} \sqrt{5-\mathrm{x}^2} \mathrm{dx} \)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app