x^2+4y^2-2x-16y+1=0
Complete the squares:
x^2-2x+1+4(y^2-4y+4)-16=0
(x-1)^2+4(y-2)^2=16
Divide through by 16:
(x-1)^2/16+(y-2)^2/4=1
This is now the equation of an ellipse in standard form: (x-h)^2/a^2+(y-k)^2/b^2=1 so a=4 (semi-major axis length) and b=2 (semi-minor axis length). The centre of the ellipse is (h,k)=(1,2). The foci can be found by using the definition that the sum of the distances between any point on the ellipse and the two foci is constant. Each focus is equidistant along the major axis from the centre. So, when y=2, x-1=±4, and x=5 and -3, distance 4 on either side of the centre. The two foci F1 and F2 are distance c on either side of the centre. The distance of x=a from F1 is a+c and from F2 is a-c so the sum is 2a. When y is distance b from the centre (vertices), the distance from each focus is the same=√(b^2+c^2), so 2√(b^2+c^2)=2a and √(b^2+c^2)=a or c=√(a^2-b^2)=√12=2√3.
So, h=1, k=2, a=4, b=2, c=2√3. Eccentricity, e=c/a=√3/2, e^2=3/4. (Note that if the ellipse had been elongated vertically, b would become the major axis and the foci would lie on this axis, so c=√(b^2-a^2).)