Let $\mathrm{C}_{1}$ be the curve obtained by the solution of differential equation $2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}^{2}-\mathrm{x}^{2}, \mathrm{x}>0$. Let the curve $C_{2}$ be the solution of $\frac{2 x y}{x^{2}-y^{2}}=\frac{d y}{d x}$. If both the curves pass through ( 1,1 ), then the area enclosed by the curves $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is equal to :

Differential Equations — JEE Main · 2021 · 16 Mar 2021 · Shift 2 Question with Solution

Let

C_{1}

be the curve obtained by the solution of differential equation

2 xy \frac{dy}{dx} = y^{2} - x^{2}, x > 0

. Let the curve

C_{2}

be the solution of

\frac{2 x y}{x ^{2} - y ^{2}} = \frac{d y}{d x}

. If both the curves pass through ( 1,1 ), then the area enclosed by the curves

C_{1}

and

C_{2}

is equal to :

Answer: (C)

Both equations are homogeneous. For

C_{1}

, putting

y = v x

2 v (v + x \frac{d v}{d x}) = v^{2} - 1 \Rightarrow \frac{2 v d v}{v ^{2} + 1} = - \frac{d x}{x}

(v^{2} + 1) x

is constant, that is

x^{2} + y^{2} = λ x

; through

(1, 1)

this gives

λ = 2

and

C_{1} : (x - 1)^{2} + y^{2} = 1

.
For

C_{2}

, putting

y = v x

x \frac{d v}{d x} = \frac{2 v}{1 - v ^{2}} - v = \frac{v ( 1 + v ^{2} )}{1 - v ^{2}} \Rightarrow (\frac{1}{v} - \frac{2 v}{1 + v ^{2}}) d v = \frac{d x}{x}

Integrating,

\frac{v}{1 + v ^{2}} = μx

, that is

\frac{y}{x ^{2} + y ^{2}} = μ

; through

(1, 1)

this gives

μ = \frac{1}{2}

and

C_{2} : x^{2} + (y - 1)^{2} = 1

.
The two circles have radius

r = 1

and centres

(1, 0)

and

(0, 1)

at distance

d = 2

, so the area common to them is

2 r^{2} cos^{- 1} \frac{d}{2 r} - \frac{d}{2} 4 r^{2} - d^{2} = 2 cos^{- 1} \frac{1}{2} - \frac{2}{2} \cdot 2 = \frac{π}{2} - 1

The area enclosed by

C_{1}

and

C_{2}

\frac{π}{2} - 1

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option