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 :
- A \(\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
- B \(\frac{1}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
- C \(\frac{1}{3}+\sqrt{5} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
- D \(\frac{2}{3}+\sqrt{5} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\)
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 \( \mathrm{A}_1=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{dx}+\int_1^{\sqrt{5}} \sqrt{5-\mathrm{x}^2} \mathrm{dx} \)…
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