The area enclosed by the closed curve C given by the differential equation
(d y)/(d x)+(x+a)/(y-2)=0, y(1)=0 is 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 R S is :

Solution

Maths sir · 5 steps

Separate the variables and you get a circle; every normal to a circle runs through its centre.

1. Separate and integrate. Write the equation as (y-2) dy + (x+a) dx = 0 and integrate both sides. This gives x² + 2ax + y² - 4y = k for some constant k, which is a circle with centre (-a, 2).
2. Fix the circle. An area of 4π means π r² = 4π, so r = 2. The circle passes through (1, 0), so (1+a)² + (0-2)² = 4. That gives a = -1, so the centre is (1, 2).

   (x-1)² + (y-2)² = 4

3. Find P and Q. Put x = 0: 1 + (y-2)² = 4, so y = 2 ± √3. This gives P(0, 2+√3) and Q(0, 2-√3).
4. Normals through centre. The normal at P joins P to (1, 2), so its slope is -√3 and its equation is y - 2 = -√3(x-1). Putting y = 0 gives R = (1 + 2/√3, 0). The normal at Q has slope √3, and in the same way it meets the x-axis at S = (1 - 2/√3, 0).
5. So. RS = 2/√3 + 2/√3 = 4/√3 = 4√3/3, which is option (a).

The idea to keep. When a differential equation separates into x² and y² terms with equal coefficients, it is a circle. For a circle, never differentiate to find a normal: just join the point to the centre.

Answer from the NTA final answer key. Solution by PrepWiser.