Dear student,

The eccentricity of an ellipse is defined as the ratio of the distance between any point in the ellipse and the focus and distance between point and a fixed line called directrix.

            e = distance between P and F/distance between P and line D (Directrix)

Formula for eccentricity :      

               Formula for calculating e is

                b² = (a²)(e²-1)    So e²- 1 = b²/a²     or e² = b²/a² +1 

                e = √(b²/a²+1)

                Eccenticity = square root of square of semi minor axis divided by squares of semi major axis

eccentricity of hyperbola = 1+b²/a² = e         ..............1

eccentricity of conjugate = 1+a²/b² = e_c²        ..................2    (e_c = eccentricity of conjugate hyperbola)

from these two equatns , e_c  = e/(e²-1)^1/2            ......................3

f(f(e)) = e                          (n = 2 even)

f(f(f(e))) = e/(e²-1)^1/2       (n = 3 odd)

now , if n = even then

integral ede = e²/2 lim from 1 to 3

                  =4                                   (A)

if n = odd then

 integral  ede/(e²-1)^1/2  lim 1 to 3

      = (e²-1)^1/2          lim 1 to 3

       =2root2                                     (D)

hence option A,D are correct ....

approve if u like my ans