The abscissa of vertex c is 4, which passes through point (0,).

∴y=a(x-4)2+k……①

In addition, the symmetry axis is a straight line x=4, and the line segment length of the image cut on the X axis is 6.

∴A( 1,0),B(7,0)

∴0=9a+k……②

A= and k= are obtained from ① ② solution.

The analytic formula of the quadratic function is: y = (x-4) 2-

(2) Point A and Point B are symmetrical about the straight line x=4.

∴PA=PB

∴PA+PD=PB+PD≥DB

When the point P is on the line segment DB, PA+PD takes the minimum value.

The intersection of db and symmetry axis is point p.

Let the straight line x=4 and the x axis intersect at point m.

Pm od, ∴∠BPM=∠BDO and ∠PBM=∠DBO.

∴△BPM∽△BDO

∴ ∴

The coordinate of point p is (4,)

(3) From (1), we can know that point C(4,),

Am = 3, ∴△AMC, cot∠ACM= in Rt,

∴∠acm=60o,∵ac=bc,∴∠acb= 120o

(1) when the point q is above the x axis, q is qn ⊥ and the x axis is n.

If AB=BQ, it is given by △ABC∽△ABQ.

BQ=6, ∠ABQ= 120o, then ∠QBN=60o.

∴QN=3，BN=3，ON= 10，

At this time Q( 10,),

If AB=AQ, Q(-2,) is known from symmetry.

② When the Q point is below the X axis, △QAB is △ACB.

At this time, the coordinate of point q is (4,),

After investigation, the points (10,) and (-2,) are all on the parabola.

To sum up, there is such a point Q, that is, △QAB∽△ABC.

The coordinates of point Q are (10,) or (-2,) or (4).