Let the maximum area of the triangle that can be inscribed in the ellipse x^2/a^2 + y^2/4 = 1, a >2

Question

Let the maximum area of the triangle that can be inscribed in the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{4}=1\), 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 ellispe is :

(A) √3/2

(B) 1/2

(C) 1/√2

(D) √3/4

Answer 1

Correct option is (A) √3/2