Question 628086
ellipse with foci at (±4, 0) and a minor axis of length 6

F = focal distance to center (0,0) = 4

j = semi major axis

n = semi minor axis = 6/2 = 3

Solve for j:

F = sqrt(j^2 - n^2)

4 = sqrt (j^2 - 3^2)

16 = j^2 - 9

j^2 = 16 + 9 = 25

j = 5

Rewriting ellipse equation in standard form

(x/a)^2 + (y/b)^2 = 1

a = 5 , b = 3

(x/5)^2 + (y/3)^2 = 1

x^2/25 + y^2/9 = 1

http://www.wolframalpha.com/input/?i=%28x%2F5%29%5E2+%2B+%28y%2F3%29%5E2+%3D+1