three

four

X and BC intersect at point D, and the ordinate of point D is -3.

The coordinate of point d is (4, -3). (2 points)

(2) On the parabola ∫ A (6,0), substitute the expression of parabola to get a=

three

eight

∴y=

three

eight

x2-

nine

four

X.(4 points)

(3) The intersection point P 1 between the parabola symmetry axis and the X axis satisfies the condition.

∫OA∨CB，

∴∠P 1OM=∠CDO.

∠∠op 1M =∠DCO = 90，

∴ rt △ p 1om ∽ rt △ CDO。 (6 points)

∵ Symmetry axis of parabola x=3,

∴ The coordinate of point P 1 is p1(3,0). (7 points)

When the vertical intersection o is OD, the symmetry axis of parabola is at point P2.

The symmetry axis is parallel to the y axis,

∴∠P2MO=∠DOC.

∠∠P2OM =∠DCO = 90，

∴ rt△ p2mo ∽ rt△ doc。 (8 points)

Point P2 also meets the requirements ∠ OP2m = ∠ ODC.

∴p 1o=co=3,∠p2p 1o=∠dco=90，

∴ rt △ p2p 1o ≌ rt △ dco。 (9 points)

∴P 1P2=CD=4.

P2 point is located in the first quadrant,

The coordinate of point P2 is P2 (3 3,4),

∴ There are two qualified points P, namely P1(3,0) and P2 (3 3,4). ( 1 1).

This topic examines the relationship between function properties and coordinates, and finally explores the existence of points, the form of geometric figures and the properties of right triangles.

Hope to adopt. Thank you.