66Let us denote the semi-major axis of En as an, the semi-minor axis as bn and the eccentricity as en.
According to the given conditions,
b_n=a_{n-1} and a_ne_n=b_{n-1}
Now,
b_n^2=a_n^2(1-e_n^2)
from where we obtain a recurrence relation
e_n^2=\dfrac{1-e_{n-1}^2}{2-e_{n-1}^2}
From this relation the third part easily follows. The locus of (en2, en-12) is
x=\dfrac{1-y}{2-y}\quad \Rightarrow \ y = \dfrac{2x-1}{x-1}
For the first part, if en is independent of n, it must be a fixed point of the sequence {en}. If that is denoted as e, then
e^2=\dfrac{1-e^2}{2-e^2}
Solving for e, we get
e=\dfrac{\sqrt{5}-1}{2}
Hence,
e_n=e_{n-3}=\dfrac{\sqrt{5}-1}{2}
Now can you do the 2nd one..we require a diameter in this case.
