∫∠BCA = 90 degrees,

∴∠bcd=∠cao=90-∠ACO；

BC = AC，∠ BDC = ∠ AOC = 90，

∴△bdc≌△coa；

∴CD=OA=2,BD=OC= 1，

∴B(-3, 1).

(2) Because the parabola passes through point B, there are:

2a×9+(-3)？ a- = 1，

Solve a =；;

∴y= x2+ x-。

(3) Let the translated triangle be △ a ′ b ′ c ′;

When y=2, x2+ x- =2,

The solution is x=3 (negative values are discarded);

∴a′(3,2),c′(2,0)；

∴ The translation process takes 3÷ 1=3 seconds;

S scan =S△ABC+S? AA′C′C

= ×( )2+3×2=8.5 (square unit).

(4)① If AC is a right-angled edge, C is a right-angled vertex;

Let the straight line BC and the parabola y= x2+ x- intersect at P 1,

The analytical formula of line BC is y =-x-； x-;

It is not difficult to get P 1( 1,-1), when CP1= AC;

∴△ACP 1 is an isosceles right triangle;

(2) If AC is a right-angled edge, point A is a right-angled vertex;

Let A be AF∨BC, and the cross parabola y= x2+ x- at P2, it is easy to get the analytical formula of straight line AF as Y =-x+2;

It is not difficult to get P2 (,) or (,).

At this time, AP2≠AC,

∴△ACP2 is not an isosceles right triangle;

∴ There is a qualified point P: P( 1,-1).