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JEE MainMathematicsArea Under Curves

The area of the region S = (x, y) : y^2 4 x 8 y^2 + 4 is:

Options

  1. A- 2 3
  2. B4 - 4 3
  3. C2 + 4 3
  4. D2 - 4 3

Correct answer

D. 2 - 4 3

Step-by-step solution

The given region is bounded by the curves x = y^2 4 and x = 8 y^2 + 4 . To find the points of intersection, we equate the two expressions for x : y^2 4 = 8 y^2 + 4 y^4 + 4y^2 - 32 = 0 (y^2 + 8)(y^2 - 4) = 0 y = 2 (since y^2 0 ) The area of the region S is given by: A = _ -2 ² ( 8 y^2 + 4 - y^2 4 ) dy Since the integrand is an even function of y , we can write: A = 2 ₀² ( 8 y^2 + 4 - y^2 4 ) dy = 2 [ 8 ( 1 2 ⁻¹ ( y 2 ) ) - y^3 12 ]₀² = 2 [ 4 ⁻¹(1) - 8 12 ] = 2 ( 4 ( 4 ) - 2 3 ) = 2 - 4 3 Answer: 2 - 4 3

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