Question 198908
4x^2 + 16y^2 = 16

is

x^2/4 + y^2/1 = 1

or

x^2/2^2 + y^2/1^2 = 1



This is 

X^2/a^2 + y^2/b^2 = 1 where a=2 and b=1.



b^2 = a^2(1-e^2) for the eccentricity e and the foci +/- ae


1 = 4 (1-e^2); ¼ = 1-e^2 giving e = (3^0.5)/2


ae = 3^0.5




Therefore the foci are (-3^0.5, 0) and (3^0.5, 0) which are approximately (-1.73205, 0) and (1.73205, 0)