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JEE Main20198 Apr 2019Evening ShiftMathematicsApplication of DerivativesActual

Given that the slope of the tangent to a curve y = y ( x ) at any point x , y is 2 y x 2 . If the curve passes through the centre of the circle x 2 + y 2 - 2 x - 2 y = 0 , then its equation is

Options

  1. Ax 2 log e ⁡ | y | = - 2 ( x - 1 )
  2. Bx log e ⁡ | y | = 2 ( x - 1 )
  3. Cx log e ⁡ | y | = - 2 ( x - 1 )
  4. Dx log e ⁡ | y | = x - 1

Correct answer

B. x log e ⁡ | y | = 2 ( x - 1 )

Step-by-step solution

Slope of tangent, d y d x = 2 y x 2 ⇒ d y y = 2 x 2 d x Integrating both sides, we get, ln ⁡ y = - 2 x + c ∵ it passes through centre 1 ,   1 of given circle, ⇒ 0 = - 2 + c ⇒ c = 2 ∴ solution ln ⁡ y = - 2 x + 2 ⇒ x ln ⁡ y = - 2 + 2 x ⇒ x ln ⁡ y = 2 x - 1

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