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MHT CET202620 April 2026Evening ShiftMathematicsApplication of DerivativesActual

The line x + y = 0 touches the curve y^2 = ax^3 + b at (1, -1) then values of a and b respectively are ...........

Options

  1. A1 2 , 2 5
  2. B1 3 , 2 3
  3. C2 3 , 1 3
  4. D2 5 , 1 2

Correct answer

C. 2 3 , 1 3

Step-by-step solution

Since the point (1, -1) lies on the curve y^2 = ax^3 + b , substituting x = 1 and y = -1 gives: (-1)^2 = a(1)^3 + b a + b = 1 Differentiating the equation of the curve with respect to x gives: 2y dy dx = 3ax^2 dy dx = 3ax^2 2y The slope of the tangent at (1, -1) is: . dy dx |_ (1, -1) = 3a(1)^2 2(-1) = - 3a 2 The given tangent line is x + y = 0 , which has a slope of -1 . Equating the slopes gives: - 3a 2 = -1 a = 2 3 Substituting the value of a into a + b = 1 gives: 2 3 + b = 1 b = 1 3 Answer: 2 3 , 1 3

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