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MHT CET202615 April 2026Evening ShiftMathematicsApplication of DerivativesActual

The point on the curve 9y^2 = x^3 where the normal to the curve makes equal intercepts with the co-ordinate axes is

Options

  1. A(-4, 8 3 )
  2. B(4, 8 3 )
  3. C(-4, -8 3 )
  4. D(-4, 3 8 )

Correct answer

B. (4, 8 3 )

Step-by-step solution

The equation of the given curve is 9y^2 = x^3 . Differentiating both sides with respect to x , we get: 18y dy dx = 3x^2 dy dx = 3x^2 18y = x^2 6y The slope of the normal to the curve is given by - dx dy : Slope of normal = - 6y x^2 Since the normal makes equal intercepts with the coordinate axes, its slope must be -1 . - 6y x^2 = -1 x^2 = 6y y = x^2 6 Substituting y = x^2 6 into the equation of the curve 9y^2 = x^3 , we get: 9 ( x^2 6 )^2 = x^3 9 x^4 36 = x^3 x^4 4 = x^3 x^4 - 4x^3 = 0 x^3(x - 4) = 0 For a non-zero

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