(2) Out of 3 points. If f crosses, the vertical line is 1 point and the vertical line in BF is 1 point. Find the center of the circle and draw ⊙P to 1 minute.

(Note: No penalty will be deducted for drawing vertical line PF without ruler)

(3) If the intersection point P is PD⊥y axis is D, then PD = X BD = 3-Y, 6 points.

PB=PF=y

△ BDP is a right triangle,

∴ Pb 2 = BD 2+PD 2 7 points

That is, ︱ y ︳ 2 = ︱ x ︳ 2+︱ 3-y ︳ 2.

That is, y 2 = x 2+(3-y) 2.

The functional relationship between y and x is y = 1/6 x 2+3/28 points.

(4) existence

The solution 1: ∵⊙ P is tangent to the X axis at point F and tangent to the straight line L at point B.

∴ AB = AF 9 points

∵AB^2=OA^2+OB^2=5^2

∴AF^2=5^2

∵AF=︱x+4︳

(x+4) 2 = 5 2 10 point

∴x= 1 or x =-9+0 1

Substitute x= 1 or x=-9 into Y = 1/6 x 2+3/2 to get y= 5/3 or y= 15.

∴ The coordinates of point P are (1, 5/3) or (-9, 15) 12 minutes.