1) parabola y=ax? -4ax+b( 1, 0) at point a and replace with:

a-4a+b=0，b=3a

y=ax？ -4ax+3a=a(x- 1)(x-3)

So: Point B is (3,0)

So: AB=2

S triangle ABC=AB* Distance from point C to X axis /2=3

So: 2*y=6, y=3.

So: point C is (0,3)

Substitution: 0+0+3a=3.

Solution: a= 1

So: the parabola is y=x? -4x+3

2)

The point F(m, 2m-5) is on the straight line y=2x-5.

The intersection of parabolas in the first quadrant f (4,3)

CF straight line: y=3

AF straight line: y=x- 1

Circle k to CF and AF all have only one common point, which is the tangent of circle K.

Distance from point K(k, 0) to tangent d=R=3.

So: d=|k-0- 1|/√2=R=3.

Solution: k= 1-3√2.

So: point K is (1-3 √ 2,0).

3)

PM=PA=PC

Then points m, a and c are all on the circle p.

So: Point P is on the middle vertical line of AC.

Ac straight line y=-3x+3

Ac middle vertical y=x/3+4/3.

At the same time as the parabola, the abscissa of point P x =( 13+√ 109)/6.

Because: Point P is on the vertical line of AM.

Therefore, the abscissa of point m satisfies: (m+1)/2 = (13+√109)/6.

Solution: m=( 10+√ 109)/3.

So the point m is ((10+√109)/3,0).