If tangents are drawn to the ellipse x 2 9 + y 2 5 = 1 at the ends of latus recta, then the area of the quadrilateral thus formed is

Step-by-step solution

The given equation of the ellipse is, x 2 9 + y 2 5 = 1 The eccentricity of the ellipse is, e = 1 - 5 9 = 2 3 The coordinates of the endpoints of latus rectum are L 2 , 5 3 , M - 2 , 5 3 , M ' - 2 , - 5 3 and L ' 2 , - 5 3 The equations of tangents at these points are, 2   x + 3   y - 9 = 0 - 2   x + 3   y - 9 = 0 2   x + 3   y + 9 = 0 - 2   x + 3   y + 9 = 0 By solving the above equations, A 0 , 3 B - 9 2 , 0 C 0 , - 3 D 9 2 , 0 And A C = ( 3 + 3 ) 2 = 6 B D = 9 2 +