F(c, 0), let P(x, y) be op * FP = (√ 3- 1) * c 2, and the area of △OFP is 2√3.

You can get the p coordinate (√ 3c, 4 √ 3/c), and OM = (c, 4/c+ 1).

Substitute the m coordinate into the elliptic equation. Eliminate a, and finally get the quadratic equation about b 2 (taking c as a constant).

The equation must guarantee: △ > =0,b^2>； 0。 Then find the range of c: R.

So when c=2, |op| is the smallest.

At this time, b 2 = 12 and a 2 = 16.

So the ellipse: x2/16+y2/12 =1.