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NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice

The slope of normal at any point P of a curve (lying in the first quadrant) is reciprocal of twice the product of the abscissa and the ordinate of point P . Then, the equation of the curve is (where, c is an arbitrary constant)

Options

  1. Ay 2 = x + c
  2. By = c e − x 2
  3. Cy = c e - x
  4. Dy 2 = ln ⁡ x + c

Correct answer

B. y = c e − x 2

Step-by-step solution

Slope of normal at P = - 1 d y d x P ⇒ 1 2 x y = - 1 d y d x or d y d x = - 2 x y ⇒ ∫ d y y = ∫ - 2 x d x ⇒ ln ⁡ y = - x 2 + k ⇒ y = e − x 2 + k = c e − x 2

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