Let the length of a latus rectum of an ellipse x^2 a^2 + y^2 b^2 =1 be 10 . If its eccentricity is the minimum value of the function f( t )= t ^2+ t + 11 12 , t R , then a ^2+ b ^2 is equal to :

Step-by-step solution

Length of LR = 2 b^2 a =10 5 a = b ^2 ...(1) aligned & f ( t )= t ^2+ t + 11 12 & df ( t ) dt =2 t +1=0 t = -1 2 & Min value of f ( t )= ( -1 2 )^2+ ( -1 2 )+ 11 12 & = 1 4 -1 2 + 11 12 = 3-6+11 12 = 8 12 = 2 3 = e & e ^2= 1- b ^2 a ^2 4 9 = 1- b ^2 a ^2 aligned b ^2 a ^2 = 1-4 a = 5 a b ^2= 5 a ^2 a ...(2) aligned & From (1) & (2) & 5 a = 5 a ^2 a a =9, ~b = 45 =3 5 & a 2+ b 2=81+45=126 aligned