An isosceles trapezium is inscribed in the ellipse x^2 a^2 + y^2 9 = 1 , where a > 3 . One of the parallel sides of the trapezium lies along the major axis of the ellipse. If the maximum possible area of this trapezium is 27 3 2 , then the length of the latus rectum of the ellipse is

Step-by-step solution

Let the vertices of the trapezium on the major axis be A(a, 0) and A'(-a, 0) . By symmetry, the other two vertices can be taken as P(a , 3 ) and Q(-a , 3 ) for some eccentric angle (0, 2 ) . The lengths of the parallel sides are 2a and 2a , and the distance between them is 3 . The area of the trapezium is A = 1 2 (2a + 2a )(3 ) = 3a(1 + ) . To maximize the area, we set dA d = 0 . dA d = 3a(- ^2 + + ^2 ) = 3a(2 ^2 + - 1) = 0 . (2 - 1)( + 1) = 0 = 1 2 = 3 . The maximum area is A_ = 3a (1 + 1 2 ) ( 3 2 ) = 9 3 4 a . G