Let the maximum area of the triangle that can be inscribed in the ellipse x 2 a 2 + y 2 4 = 1 , a > 2 , having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y -axis, be 6 3 . Then the eccentricity of the ellipse is:

Step-by-step solution

Plotting the diagram as per given information we get, Now finding area of the triangle we get, A = 1 2 a 1 - cos θ 4 sin θ A = 2 a 1 - cosθ sin θ Differentiating to get maxima and minima we get, dA d θ = 2 a sin 2 θ + cos θ - cos 2 θ d A   d θ = 0 ⇒ 1 + cos θ - 2 cos 2 θ = 0 cos θ = 1 Reject OR cos θ = - 1 2 ⇒ θ = 2 π 3 d 2   A   d θ 2 = 2 a 2 sin 2 θ - sin θ d 2   A   d θ 2 < 0 fo