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MHT CET202616 April 2026Evening ShiftMathematicsApplication of DerivativesActual

The equation of tangent to the curves x = 1 - 3t^2 and y = t - 3t^3 at the point (-2, 2) is...

Options

  1. A4x + 3y + 2 = 0
  2. B4x - 3y + 2 = 0
  3. C3x + 4y + 2 = 0
  4. D3x - 4y + 2 = 0

Correct answer

A. 4x + 3y + 2 = 0

Step-by-step solution

Given x = 1 - 3t^2 and y = t - 3t^3 . At the point (-2, 2) , we have 1 - 3t^2 = -2 3t^2 = 3 t = 1 . Also t - 3t^3 = 2 . For t = -1 , -1 - 3(-1)^3 = 2 , which satisfies the equation. Thus, t = -1 . Differentiating x and y with respect to t : dx dt = -6t dy dt = 1 - 9t^2 The slope of the tangent is dy dx = dy dt dx dt = 1 - 9t^2 -6t . At t = -1 , dy dx = 1 - 9(-1)^2 -6(-1) = -8 6 = - 4 3 . The equation of the tangent at (-2, 2) is: y - 2 = - 4 3 (x - (-2)) 3(y - 2) = -4(x + 2) 3y - 6 = -4x - 8 4x + 3y + 2 = 0 Answer:

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