JEE Main 2021 · 18 March · morning · Single correct · +4 / −1

The differential equation satisfied by the system of parabolas
y² = 4a(x + a) is :

  1. (a) y( dy/dx )² - 2x( dy/dx ) - y = 0
  2. (b) y( dy/dx )² - 2x( dy/dx ) + y = 0
  3. (c) y( dy/dx )² + 2x( dy/dx ) - y = 0
  4. (d) y( dy/dx ) + 2x( dy/dx ) - y = 0
Answer and solution

Answer (c) y( dy/dx )² + 2x( dy/dx ) - y = 0

Solution

Maths sir · 4 steps

One arbitrary constant means one differentiation: find a from y' and put it back in.

  1. Differentiate once. The family has one constant a, so you need one derivative. Write y² = 4ax + 4a² and differentiate with respect to x: 2ydy/dx = 4a. Let p = dy/dx, so a = yp/2.
  2. Eliminate a. Put a = yp/2 back into y² = 4ax + 4a². This gives y² = 2xyp + y²p².

    y² = 2xy p + y²p² ⇒ y = 2xp + yp²

  3. Tidy up. Dividing by y is allowed, since y ≠ 0 for points on the curves away from the vertex. Then take every term to one side: yp² + 2xp - y = 0.
  4. So. The required equation is y(dy/dx)² + 2xdy/dx - y = 0, which is option (c).

The idea to keep. Count the arbitrary constants and differentiate that many times. Then use the derivatives to express each constant and substitute back, so no constant is left in the final equation.

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

Study Differential Equations

Asked about 200 times in JEE Main since 2020. This question is from 2021.