The distance between the foci is 2√15 or approx 7.746.
The standard equation for an ellipse is (x-h)^2/(a^2) + (y-k)^2/(b^2) = 1.
In this equation, a and b are the semi-major and semi-minor axes.
First rewrite the given equation in standard form by taking the square root of the denominators:
(x-0)^2/(8^2) + (y-0)^2/(7^2) = 1.
Note that x-0 = x and y-0 = y, but writing it this way shows where the center (h,k) is.
The center of the ellipse is (0,0).
Also a=8, and b=7.
Since a>b, a is the semi-major axis, leaving b to be the semi-minor axis.
Next, find the focus points. An ellipse with center (h,k) and major axis parallel to the x-axis has foci (h+c, k) and (h-c, k). For the given equation, this is shown as (0+c, 0) and (0-c, 0).
The equation c = √{(a^2) - (b^2)} is the distance from the foci to the center.
[Note: all of that above is the radicand; under the radical].
Next, substitute the values found for a and b into that equation:
c = √{(8^2) - (7^2)}
c = √{(64) - (49)}
c = √15
Now use this c value to find the focus points:
(0+√15, 0) and (0 - √15, 0).
Simplify:
(√15, 0) and (-√15, 0).
Finally, the eccentricity e=√{(a^2)-(b^2)}/a
e = (√15)/8 or approx 0.484.
The distance between the two focus points are equal to 2ae.
Given a=8 and e=(√15)/8,
2*(8)*((√15)/8) = 2√15 or approx 7.746.