So: a = 2c/√ 3 (a > b), e = c/a = √ 3/2.

(2) point f is (-c, 0) and point b is (0, c/√3)

The slope of FB is: c/√ 3/c =1√ 3, then the slope of BP is: -√3.

The BP equation is: y-c/√3=-√3x, then: p (c/3,0).

Because FB⊥BP, the circle passing through f, b and p has FB as its diameter.

So the center of the circle is the midpoint of FP: (-c/3,0), and the radius r=│PF│/2=2c/3.

Distance from (-c/3,0) to straight line x+√3y-√3=0:

d=│-c/3+0-√3│/√[ 1^2+(√3)^2]=(c/3+√3)/2

If a circle is tangent to a straight line, d = r.

Then: 2c/3=(c/3+√3)/2, the solution is: c=√3.

a=2c/√3=2，b=c/√3= 1

So the elliptic equation is: x 2/4+y 2 =1.