An isosceles trapezium is inscribed in the region bounded by the parabola y^2 = 4x and its latus rectum. One of the parallel sides of the trapezium is the latus rectum itself, and the other parallel side is a chord of the parabola perpendicular to its axis. If the area of this trapezium is maximum, then the length of the chord forming the second parallel side is :

Step-by-step solution

For the parabola y^2 = 4x , the latus rectum lies on the line x = 1 . Its length is 4 . Let the other parallel side be a chord along the line x = t , where 0 The endpoints of this chord are (t, 2 t ) and (t, -2 t ) , so its length is 4 t . The distance between the parallel sides (height of the trapezium) is 1 - t . The area of the trapezium is: A(t) = 1 2 (4 + 4 t )(1 - t) = 2(1 + t )(1 - t) Let u = t . Then A(u) = 2(1 + u)(1 - u^2) = 2(1 + u - u^2 - u^3) . To maximize the area, we differentiate A(u) with respect t