D should be (0, (7 √ 3)/9) (the root sign of 3/9 of 0, 7).

( 1)

∵C is the vertex, and the abscissa of C is 4.

The ∴ symmetry axis of a quadratic function is a straight line x=4.

AB = 6

∴A( 1,0) B(7，0)

∫D(0，(7√3)/9)A( 1，0) B(7，0)

∴0=a+b+(7√3)/9

0=49a+7b+(7√3)/9

Y = ... Do it yourself ... This young lady is too lazy to forget. ...

(2)

Let A be the symmetrical point of the straight line x=4 (that is, point B).

Connect DB, and its intersection with the straight line x=4 is p.

The analytical formula of easy-to-find DB is y = ... Use the undetermined coefficient method to calculate for yourself ... only provide your method. ...

Let x=4 in BD analytical formula, and find the value of y, where y is the ordinate of point P and 4 is the abscissa of point P.

That is, looking for p.

(3)

exist ...

Find a point on the parabola ... so the distance between this point and A or B is 6 ... This point is the sought point Q. ...

When the q point is on the left side of the straight line x=4

∵cos∠CAB= a certain value (too lazy to forget it)

∴cos∠AQB= has some value.

AQ=6

∴QB=2*6*cos∠AQB

Use the coordinate difference formula: the distance between A and B under the radical sign [(xa-XB) 2-(ya-Yb) 2] = 2.

Substituting the coordinates of point B and the distance of QB, you can get the coordinates of point Q.

This Q point and its symmetrical point about the straight line x=4 are the solutions.

I'm sorry ... I really don't have time to calculate ... I didn't know you didn't have time when you were in primary school ... in the third grade. ...