JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Differential Equations

The area enclosed by the closed curve C given by the differential equation

\frac{d y}{d x} + \frac{x + a}{y - 2} = 0, y (1) = 0

4 π

.

Let P and Q be the points of intersection of the curve C and the y -axis. If normals at P and Q on the curve C intersect x -axis at points R and S respectively, then the length of the line segment RS is

(A) $23$
(B) $\frac{2 3}{3}$
(C)2
(D) $\frac{4 3}{3}$ ✓ Correct

Answer: (D)

Solution

Separating variables,

(y - 2) d y = - (x + a) d x

integrates to

(x + a)^{2} + (y - 2)^{2} = 2 C

, a circle of radius

2 C

. An enclosed area of

4 π

fixes the radius at

2

, so

2 C = 4

, and passing through

(1, 0)

gives

(1 + a)^{2} + 4 = 4

, hence

a = - 1

. The circle is

(x - 1)^{2} + (y - 2)^{2} = 4

with centre

(1, 2)

, meeting the

y

-axis at

P (0, 2 + 3)

and

Q (0, 2 - 3)

. The normals at

P

and

Q

pass through the centre and cut the

x

-axis at

R (\frac{2 + 3}{3}, 0)

and

S (\frac{3 - 2}{3}, 0)

, so

R S = \frac{4}{3} = \frac{4 3}{3}

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Differential Equations

Solution

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