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JEE Main202118 Mar 2021Morning ShiftMathematicsDifferential EquationsActual

The differential equation satisfied by the system of parabolas y 2 = 4 a ( x + a ) is

Options

  1. Ay d y d x 2 - 2 x d y d x - y = 0
  2. By d y d x 2 - 2 x d y d x + y = 0
  3. Cy d y d x 2 + 2 x d y d x - y = 0
  4. Dy d y d x + 2 x d y d x - y = 0

Correct answer

C. y d y d x 2 + 2 x d y d x - y = 0

Step-by-step solution

y 2 = 4 a x + 4 a 2 differentiate with respect to x ⇒ 2 y d y d x = 4 a ⇒ a = y 2 d y d x so, required differential equation is y 2 = 4 × y 2 d y d x x + 4 y 2 d y d x 2 ⇒ y 2 d y d x 2 + 2 x y d y d x - y 2 = 0 ⇒ y d y d x 2 + 2 x d y d x - y = 0

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