Find an equation of the ellipse having a major axis of length 8 and foci at $(-1,-5)$ and $(-5,-5)$.

Step 1 ：The equation of an ellipse in standard form is given by: \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\) where \((h,k)\) is the center of the ellipse, \(a\) is the length of the semi-major axis, and \(b\) is the length of the semi-minor axis.

Step 2 ：The foci of the ellipse are located at \((h±c,k)\), where \(c\) is the distance from the center of the ellipse to each focus.

Step 3 ：Given that the foci are at \((-1,-5)\) and \((-5,-5)\), we can find the center of the ellipse by finding the midpoint of the segment connecting the foci. The midpoint formula is: \(((x₁+x₂)/2, (y₁+y₂)/2)\)

Step 4 ：So the center of the ellipse is: \(((-1-5)/2, (-5-5)/2) = (-3,-5)\)

Step 5 ：The distance from the center to each focus (\(c\)) is the absolute value of the difference in x-coordinates of the center and one of the foci, which is \(|(-3)-(-1)| = 2\).

Step 6 ：The length of the major axis is given as 8, so \(a = 8/2 = 4\).

Step 7 ：We can find \(b\) using the relationship \(a^2 = b^2 + c^2\) in an ellipse. Solving for \(b\) gives us: \(b = \sqrt{a^2 - c^2} = \sqrt{4^2 - 2^2} = \sqrt{12} = 2\sqrt{3}\)

Step 8 ：So the equation of the ellipse is: \((x+3)^2/4^2 + (y+5)^2/(2\sqrt{3})^2 = 1\)

Step 9 ：Simplifying gives: \((x+3)^2/16 + (y+5)^2/12 = 1\)

Step 10 ：This is the equation of the ellipse in standard form. So, the final answer is \(\boxed{(x+3)^2/16 + (y+5)^2/12 = 1}\)