The maximum area in square units of an isosceles triangle inscribed in an ellipses x² a² + y² b² =1 with its vertex at one end of the major axis is

Step-by-step solution

Let A B C be an isosceles triangle inscribed in the ellipse x² a² + y² b² =1 . Let coordinate of A be (a, 0) . Area of A B C, A= 1 2 A D B C aligned &= 1 2 (a+x) 2 y=y(a+x) &= b² (1- x² a² ) (a+x)= b a a²-x² (a+x) aligned d A d x = b a [(a+x) 1 2 -2 x a²-x² + a²-x² ]= b a [ -x(a+x)+ (a²-x² ) a²-x² ] Put d A d x =0 2 x²+a x-a²=0 2 x²+2 a x-a x-a²=0 (2 x-a)(x+a)=0 x= a 2 ,-a Since, d² A d x² < 0 , for x= a 2 aligned Maximum area &= b a (a+ a 2 ) a²- a² 4 &= b a 3 2 (a) 3 a 2 = 3 3 a b 4 sq units aligned