24. Solution: (1) According to the meaning of the question, let the analytical formula of parabola be y=a(x- 1)2- 1, then 0=a(0- 1)2- 1.

∴ a =1..................................... (2 points)

∴y=(x- 1)2- 1, that is, y = x2-2x ............................ (3 points).

(2) When y=0, x2-2x=0 gives x=0 or x=2.

∴ A (2 2,0) .................................. (4 points)

B( 1，- 1)，O(0，0)，

∴OB2=2，AB2=2，OA2=4。

∴OB2 + AB2 = OA2

∴∠OBA = 900，OB=BA。

∴△OBA is an isosceles right triangle.

Form .......... (6 points)

(3) As shown in the figure, C is CE∨BO and CF∨AB, with a score of 0.

The parabola should not intersect with point F at point E, but after point F..

If the FD⊥X axis is on the D axis, then ∠ECF=900, EC=CF, FD=CD.

∴△ECF is an isosceles right triangle ........................ (7 points).

Let FD = m > 0, then CD = m, OD =1+m.

∴ F( 1+m, m) ................................... (8 points)

∴ m =( 1+m)2-2( 1+m)，

That is m2-m- 1=0. The solution is m=

∵m>0，∴m=。

∴F()。

∵ points e and f are symmetrical about the straight line x= 1,

∴ e = () .......................... (10)