(2)
Note that F1 and F2 are endpoints of the diameter of the circle....so triangle F1BF2 is a right triangle inscribed in the circle....
And note that sqrt (25) = 5 = a
And 2a = 10
And the focal length = the radius of the circle = sqrt (a^2 - b^2) = sqrt (25 -16) = sqrt (9) = 3
So F1F2 = 2(3) = 6
And since triangle F1BF2 is right with F1F2 the hypotenuse and BF2 a leg = 5 then BF1 =
sqrt (6^2 - 5^2) =sqrt (36 -25) = sqrt (11)
Call F1A = y
And AB =x
Since triangle F1BF2 is right, then so is triangle F1AB with ABF1 being a right angle
And the sum of the distances F1A and F2A = 2a = 10
So we have the following system
(F1A)^2 - (AB)^2 = (BF1)^2
F1A + F2A = 6
So we have that
y^2 - x^2 =11 (1)
y + (5 + x) = 10 ⇒ x + y = 5 ⇒ y = 5 - x (2)
Subbing (2) into (1) we have that
(5 - x)^2 - x^2 = 11
25 - 10x = 11
14 = 10x
7 = 5x
x =7/5
And y = 5 - (7/5) = 25/5 -7/5 =18 / 5
So....the cosine of F1AF2 =
AB / AF1 =
x / y =
(7/5) / (18/5) =
7 / 18