Too lazy to name pfy

The two focuses of an ellipse are F 1(-p, p) and F2(p, p).

∴ The focal length of the ellipse 2c=|F 1F2|=6.

∵P On the ellipse, the maximum area of △PF 1F2 is 12.

∴ When the point P falls on the endpoint of the short axis, the area of △PF 1F2 is s =12×| f1F2 |× 5 =12.

That is, p5= 12 and 5=3, so 4 = 52+C2 = 5.

So the eccentricity of the ellipse is seven =c four =p5.

So the answer is: p5.