Let the length of a latus rectum of an ellipse x2a2 + y2b2 = 1 be 10. If its eccentricity is the minimum value of the function f(t) = t2 + t + 1112, t R, then a2 + b2 is equal to :
- 115
- 120
- 125
- 126
Answer
(4) 126
Solution
f(t) = (t + 12 )2 + 1112 - 14 minimum value = 23
e = 23 (taking a > b, major axis along x-axis)
b2 = a2(1 - e2) = 5a29
Latus rectum: 2b2a = 10 2a 5a29 = 10 a = 9
a2 = 81, b2 = 45
a2 + b2 = 126
Answer: (4)