Question

Find the equation of an ellipse that have eccentricity 2/5 and foci (-1, 3) and (3, 3) .

Answer 1

distance from (-1,3) to (3,3) is 4, so 2c = 4
c = 2
for your ellipse , e = c/a
2/5 = 2/a
a = 5
we also know that a^2 = b^2 + c^2
25 = b^2 + 4
b^2 = √21
the centre is midway between the foci, so the centre is (1,3)

equation:
(x-1)^2 / 25 + (y-3)^2 / 21 = 1

verification:
https://www.wolframalpha.com/input/?i=plot+%28x-1%29%5E2+%2F+25+%2B+%28y-3%29%5E2+%2F+21+%3D+1

notice the a^2 is close to b^2, so it looks almost circular