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JEE Main20264 April 2026Evening ShiftMathematicsArea Under CurvesActual

The area of the region bounded by the curves x+3y^2=0 and x+4y^2=1 is equal to:

Options

  1. A1 3
  2. B2 3
  3. C4 3
  4. D5 3

Correct answer

C. 4 3

Step-by-step solution

The equations of the given curves are x = -3y^2 and x = 1 - 4y^2 . To find the points of intersection, equate the values of x : -3y^2 = 1 - 4y^2 y^2 = 1 y = 1 The region is bounded between y = -1 and y = 1 . In this interval, 1 - 4y^2 -3y^2 . The required area A is given by: A = _ -1 ¹ ( (1 - 4y^2) - (-3y^2) ) dy A = _ -1 ¹ (1 - y^2) dy Since 1 - y^2 is an even function: A = 2 ₀¹ (1 - y^2) dy A = 2 [ y - y^3 3 ]₀¹ A = 2 ( 1 - 1 3 ) = 2 ( 2 3 ) = 4 3 Answer: 4 3

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