The abscissa of the vertex m is 1 and the symmetry axis is x=-b/2a= 1.

When the two equations are combined, A = 1 and B =-2 can be solved.

∴ Parabolic equation is y=x? -2x-3

(2) Vertex coordinates can be easily obtained from parabola as M( 1, -4).

Let the coordinates of p and q be p (p, n) and q (1, q).

If point p is on a parabola, then there is n=p? -2p-3 ( 1)

Because AQ⊥PQ, there is AQ 2+PQ 2 = AP 2.

That is, [2 2+Q2]+[(p-1) 2+(n-q) 2] = [(p+1) 2+N2] (2).

Let the intersection of QM and X axis be n, then AN=2, MN=4, AM=2√5.

∵∠PAQ=∠AMQ,∴sin∠PAQ=sin∠AMQ

That is pq/AP = an/am =1√ 5, that is, [(p-1) 2+(n-q) 2]/[(p+1) 2+N2] = 65438+.

The values of simultaneous (1)(2)(3), n, p, q P and q can be obtained by solving equations.

This is a quartic equation, which is quite troublesome to solve. I get an approximate value by mirror image method.

There are four groups of solutions, namely:

p=-2.45，n=7.90，q = 6.90

p=2.45，n=- 1.90，q =-2.90；

p=4.00，n=5.00，q = 6.00

p=0.00，n=-3.00，q =-2.00；

(3) Let the coordinate of point R be R(r, 0), and the parabola rotates around r 180.

A new parabola c 1: y =-a (x-h) 2+k is obtained.

A (- 1, 0), B (3 3,0), M (1, 4) and N (1, 0) can be easily obtained from the known results.

The coordinate of the new symmetry point E is E(2r+ 1, 0).

Then MN=4, MB=MA=2√5, BE=|2r-2|, NE=|2r|.

me=√(mn^2+ne^2)=√[4^2+4r^2]=2√[4+r^2]

∫∠BME = 45, applying cosine theorem to △BME, we can get

BE^2=MB^2+ME^2-2MB*ME*cos∠BME

That is, (2r-2) 2 = (2 √ 5) 2+4 (4+R2)-2 * 2 √ 5 * 2 √ [4+R2] * COS45.

You can get 3r 2- 16r- 12 = 0 after sorting.

The solution can be obtained with r=-2/3 or r=6.

That is, the coordinates of the R point are R (-2/3,0) or R (6,0).