Area Under the Curve
Parabola Dividing Circle
nta_pyq_2024_apr
Grade 12
Question:
The parabola $y^2=4x$ divides the area of the circle $x^2+y^2=5$ in two parts. The area of the smaller part is equal to:
$\dfrac{1}{3}+5\sin^{-1}\left(\dfrac{2}{\sqrt{5}}\right)$
$\dfrac{1}{3}+\sqrt{5}\sin^{-1}\left(\dfrac{2}{\sqrt{5}}\right)$
$\dfrac{2}{3}+5\sin^{-1}\left(\dfrac{2}{\sqrt{5}}\right)$
$\dfrac{2}{3}+\sqrt{5}\sin^{-1}\left(\dfrac{2}{\sqrt{5}}\right)$
Step-by-Step Solution
Key Concept: Area $=2A_1$ where $A_1=\int_0^1\sqrt{4x}dx+\int_1^{\sqrt{5}}\sqrt{5-x^2}dx$.
Smaller area $=\dfrac{2}{3}+5\sin^{-1}\left(\dfrac{2}{\sqrt{5}}\right)$.
Correct Answer: 3