∴OA=2，

∴OB= 12OA= 1，

∵ point b is on the positive semi-axis of the y axis,

∴ The coordinate of point B is (0,1);

C is the CD⊥x axis, and the vertical foot is d,

∵BA⊥AC,∴∠OAB+∠CAD=90，

∠ AOB = 90，∴∠ OAB+∠ OBA = 90，

∴∠CAD=∠OBA，AB=AC，∠ AOB =∠ ADC = 90，

∴△AOB≌△CDA，

∴OA=CD=2,OB=AD= 1，

∴OD=OA+AD=3, c is the point of the first quadrant,

∴ The coordinate of point C is (3,2);

(2) ∵ Both point B and point C are on parabola, y=-56x2+bx+c,

∴ replace B (0, 1) and C (3 3,2),

Get c = 1? 56×9+3b+c=2，

The solution is b = 176c = 1,

The analytical formula of parabola is y =-56x2+176x+1;

(3) There is a point P on the parabola, and △ACP is an isosceles right triangle with AC on the right, which is divided into three cases:

(i) If AC is a right-angled edge and point A is a right-angled vertex, extend BA to point P 1 so that P 1A=CA, and obtain an isosceles right-angled triangle ACP 1.

P 1 is the P 1M⊥x axis, as shown in the figure.

∫AP 1 = CA = AB，∠MAP 1=∠OAB，∠P 1MA=∠OBA=90，

∴△AMP 1≌△AOB，

∴AM=AO=2,P 1M=OB= 1，

∴OM=OA+AM=4，

∴P 1(4,-1), and the checkpoint P 1 is on the parabola y =-56x2+1;

(ii) If AC is a right-angled side and point C is a right-angled vertex, then the intersection point C is CP2⊥AC, and CP2=AC, thus obtaining an isosceles right-angled triangle ACP2.

The intersection P2 is a parallel line of the Y axis, and the intersection C is a parallel line of the X axis. The two lines intersect at point N, as shown in the figure.

It can also be proved that △ cp2n △ ABO,

∴CN=OA=2,NP2=OB= 1，

The coordinate of ∵C is (3,2),

∴ P2 (1, 3), also on the parabola y=-56x2+ 176x+ 1