Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE MainMathematicsArea Under Curves

If the area of the region (x, y) : |2y - a| a - 2x^2 is 72 (where a > 0 ), then the value of a is

Options

  1. A36
  2. B9
  3. C18
  4. D12

Correct answer

C. 18

Step-by-step solution

Given the region (x, y) : |2y - a| a - 2x^2 . This inequality can be rewritten as: -(a - 2x^2) 2y - a a - 2x^2 Adding a to all parts: 2x^2 2y 2a - 2x^2 Dividing by 2 : x^2 y a - x^2 The region is bounded by the upward-opening parabola y = x^2 and the downward-opening parabola y = a - x^2 . To find the points of intersection, we equate the two curves: x^2 = a - x^2 2x^2 = a x = a 2 The area of the region is given by the integral of the upper curve minus the lower curve: Area = _ - a/2 ^ a/2 ( (a - x^2) - x^2 ) dx Ar

Practice Area Under Curves on Quantrex Academy →

More from Area Under Curves

Passage: Consider the curve C₁ given by y = e^ -x for x [0, 10 ] , and the curve C₂ given by y = e^ -x ( x + x) for x [0, 10 ] . Let n be the total number of points of intersection 2026Passage: Consider the ellipses given by x^2 + 4y^2 = 1 and 4x^2 + y^2 = 1 . Question: If is the area of the common region that lies inside both the given ellipses, then the value o 2026The area of the region (x, y) : x^2 - 8x y -x is : 2026The area of the region (x, y) : 0 y 6 - x, y^2 4x - 3, x 0 is: 2026The area of the region R = (x, y): xy 27, 1 y x^2 is equal to: 2026The area of the region bounded by the curves x+3y^2=0 and x+4y^2=1 is equal to: 2026The area of the region (x, y): y - |x|, y |x x|, y 0 is: 2026If the area of the region bounded by 16x^2 - 9y^2 = 144 and 8x - 3y = 24 is A, then 3(A + 6 _e(3)) is equal to _______. 2026 Full Area Under Curves list All JEE Main PYQs