JEE PYQ: Area Under Curves Question 2
Question 2 - 2021 (16 Mar Shift 2)
Let $C_1$ be the curve obtained by the solution of the differential equation $2xy\frac{dy}{dx} = y^2 - x^2$, $x > 0$. Let the curve $C_2$ be the solution of $\frac{2xy}{x^2 - y^2} = \frac{dy}{dx}$. If both the curves pass through $(1, 1)$, then the area enclosed by the curves $C_1$ and $C_2$ is equal to:
(1) $\pi - 1$
(2) $\frac{\pi}{2} - 1$
(3) $\pi + 1$
(4) $\frac{\pi}{4} + 1$
Show Answer
Answer: (2) $\frac{\pi}{2} - 1$
Solution
$C_1: x^2 + y^2 - 2x = 0$, $C_2: x^2 + y^2 - 2y = 0$. Required area $= 2\int_0^1 (\sqrt{2x - x^2} - x),dx = \frac{\pi}{2} - 1$.
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