The analytical formula of the straight line is: y=-43x+4,

The analytical formula of parabola is: y = x2-4x+3;

(2)(2) If ⊙P is tangent to the straight line AB and the X axis,

Then point P is on the straight line where ∠BAO or the bisector of its outer corner is located.

① Let the bisector of abalone intersect the Y axis at D, and let D be DH⊥AB at H,

Then DH=DO=m, BD=4-m, AH=AO=3, BH=5-3=2.

In Rt△BHD, BD2=BH2+DH2.

That is, (4-m)2=m2+22,

Solution: m=32

That is d (0, 1.5)

The analytical formula of straight line AD is: y=- 12x+32,

Solve it simultaneously with the parabolic analytical formula y=x2-4x+3: {x1= 3; y 1=0，{ x2 = 12； y2=54

P ( 12，54)

② Let ∠BAO's external angle bisector intersect with Y axis at G,

Then AG⊥AD is in a, then △DOA∽△AOG, so OG=2OA=6.

That is, the analytical formula of G(0, -6) straight line DG is: y=2x-6.

Solve it simultaneously with the parabolic analytical formula y=x2-4x+3: {x1= 3; y 1=0

∴ There is a point p (12,54), which makes ∴ p tangent to the straight line AB and the X axis.

(3) Let P be the X axis of PM ⊥ in M, obviously PM is the center line of Rt△OQE, that is, oe = 2om = 2 | x |, QE=2PM.

If the point p is on the parabola x2-4x+3, then P(x, x2-4x+3), QE=2PM=2|x2-4x+3|

① when x < 0, x2-4x+3 > 0, OE=-2x, y = 2 [-2x+2 (x2-4x+3)] = 4x2-20x+12.

② when 1 < x < 3, x2-4x+3 < 0, y = 2 [2x-2 (x2-4x+3)] =-4x2+20x-12.

③ when 0 < x < 1 or x > 3, x2-4x+3 > 0, y = 2 [2x+2 (x2-4x+3)] = 4x2-12x+12.

The problem has been solved.