∵ In Rt△OAB, tan∠OAB=OBOA= 13.
∴OA= 1,
∴ Point A (- 1, 0)
By substituting the parabola y=-x2+bx+c passing through points a and b, we get:
0=? 1? b+c3=c,
∴b=2,c=3,
The expression of parabola is y=-x2+2x+3.
∫y =-x2+2x+3 =-(x- 1)2+4,
∴ The coordinate of vertex D is (1, 4);
(2) The symmetry axis of parabola is a straight line L = x= 1,
From the meaning of the question: the symmetry point of point B about straight line L is C,
∴ The coordinates of point C are (2,3), and point E (1, 3) is the midpoint of BC.
∴DE= 1,
Point d is the center of gravity of △ △PBC,
∴PD=2DE=2,
That is, PE=3,
Starting from the point p on the straight line l,
∴ The coordinate of point P is (1, 6);
(3)∫P( 1,6),D( 1,4),
∴PD=2, you can see that the parabola y=-x2+2x+3 moves up by 2 units.
The expression after parabola translation is y=-x2+2x+5.
The coordinate of point m is (m, n).
The heights of PD on both sides of △MPD and △BPD are |m- 1| and 1 respectively.
Therefore, the area of △MPD is equal to twice the area of △BPD,
| m- 1 | = 2。
Solution: m 1=- 1, m2 = 3.
The point m is on the parabola y=-x2+2x+5,
∴n 1=2,n2=2,
∴ The coordinates of point m are M 1 (- 1, 2) and M2 (3, 2) respectively.