For a conic, the eccentricity (e) between 0 and 1 will make this an ellipse.

Since the foci are located at (6,1) and (6,-11), the distance between foci is 12 units = 2c.

The eccentricity is related to half of this distance and half of the length of the major axis of the ellipse (2a).

e = c/a

We also have a^{2}-b^{2 }= c^{2} for an ellipse.

So c = 6, which makes a = 10, which makes b = 8.

Since the center of the ellipse is at (6,-5), the equation for the ellipse is (y+5)^{2}/a^{2} + (x-6)^{2}/b^{2} = 1

or (y+5)^{2}/100+(x-6)^{2}/64=1