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MHT CET202613 April 2026Evening ShiftMathematicsApplication of DerivativesActual

If the tangent to the curve 2y^3 = x^3 + ax^2 at the point (a, a) cuts off intercepts and on the coordinate axes such that ^2 + ^2 = 61 , then the value of a is

Options

  1. A61
  2. B36
  3. C30
  4. D25

Correct answer

C. 30

Step-by-step solution

Given curve: 2y^3 = x^3 + ax^2 Differentiating with respect to x : 6y^2 dy dx = 3x^2 + 2ax At the point (a, a) , the slope of the tangent is: 6a^2 dy dx = 3a^2 + 2a^2 = 5a^2 dy dx = 5 6 The equation of the tangent at (a, a) is: y - a = 5 6 (x - a) 6y - 6a = 5x - 5a 5x - 6y + a = 0 The intercepts on the coordinate axes are and . For the x -intercept , put y = 0 : 5 + a = 0 = - a 5 For the y -intercept , put x = 0 : -6 + a = 0 = a 6 Given that ^2 + ^2 = 61 : (- a 5 )^2 + ( a 6 )^2 = 61 a^2 25 + a^2 36 = 61 a^2 ( 36 +

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