(2) As shown in the figure, let the PF⊥x axis be at point F,

The coordinates of point P are (2,-1), ∴PF= 1.

∵P is the vertex of the parabola, and A and B are the intersections of the parabola and the X axis.

∴PA=PB

∫△PAB is an isosceles right triangle.

∴PF is the centerline of the bottom of△△ PAB.

∴AB=2PF=2

From a(x-2)2-2=0, x 1=2+), x2=2-)

∴A(2-),0),B(2+),0)

∴AB=(2+))-(2-))=2)，

∴ 2) = 2-①, and the solution is a= 1 (after investigation, a= 1 is consistent with equation ① and the meaning of the question).

∴a= 1, and the analytical formula of the function is y = (x-2) 2- 1.

Let x=0 get y = 3, and the coordinate of point C is (0,3).

The coordinate of b is (3,0), and the coordinate of p is (2, 1).

The analytical formula of ∴ linear BP is y=x-3.

Substitute x=0 into y=x-3, y=-3,

The coordinate of point d is (0, -3).

To sum up, A = 1, C (0 0,3), D (0 0,3).

(3)① As shown in the figure, E(0, b), if 0

∫A( 1，0)，C(0，3)

∴OA= 1,OC=3

∴S△AOC=

As can be seen from the meaning of the question, △CEG∽△AOC

∴ = )2

∴S÷ = 2÷32

∴S= 2

② When b = 0, S=

③ When- 1

④ When b=- 1, S=

⑤ When -3