(2) The known point p is the moving point of the image of the quadratic function y=-x2+3x on the right side of the Y axis. Translate the straight line y=-2x along the Y axis, and intersect the X axis and the Y axis at C and D respectively. If the △PCD at the right angle of CD is similar to △OCD, the coordinates of point P are (2,2), (1.

Solution:

Parabola: y=-x? +3 times

The symmetry axis is x=3/2.

Straight line: y=-2x

Coordinates of intersection b (3/2, -3)

(2) The other right angle is perpendicular to DC when crossing point C, and perpendicular to DC when crossing point D.

Let the coordinates of point C be: (c, 0) (c >; 0)

Then the coordinates of point D are: (0, 2c)

The slope of DC is -2 degrees.

Then the slope of PC is 1/2, and through point C, the PC equation is:

y-0= 1/2(x-c)

That is, y=x/2-c/2.

The intersection with the parabola is:

x/2-c/2=-x？ +3 times

2x？ - 5 x-c=0

x =(5-√( 25+8c))/4 & lt； 0 is on the left side of the y axis,

Or x = (5+√ (25+8c))/4 >; 0

y=(5+√(25+8c) -4c)/8

The coordinates of point P are ((5+√(25+8c))/4, (5+√(25+8c) -4c)/8).

| PC |: | DC | = | OD |: | OC| = 2 or | PC |: | DC | = | OC |: | OD | = 1: 2

|DC|=√5c

|PC|=√((((5+√(25+8c))/4-c)？ +((5+√(25+8c) -4c)/8)？ )

=√5/2* (5+√(25+8c) -4c)/4

=√5 (5+√(25+8c) -4c)/8

C= 13/25，P( 13/5，27/50) from |PC|=2|DC|。

c = 1 1/8 from | PC | = | DC |/2，P ( 1 1/4， 16/ 1 1)。

The slope of PD is 1/2, passing through point D.

y-2c= 1/2(x-0)

That is y=x/2+2c.

The intersection with the parabola is:

x/2+2c=-x？ +3 times

2x？ -5x+4c=0

When 25-32c=0, that is, c=25/32, PD is tangent to the parabola and has an intersection point.

At 25-32, at 25-32c, there is no intersection between PD and parabola.

When 25-32c >: 0, that is, C.