In rectangular coordinate system, parabola y=ax? +bx+c intersects with X axis at point A and point B (point A is to the left of point B), and intersects with Y axis at point C, point A (-3,0) and point C (0,3). The parabola symmetry axis is x =-2 (1). If p is a point on the line segment ac, let △abp and △bpc be areas.

Solution: (1) According to the meaning of the question

Axis of symmetry x=-2

Then the coordinate of point B is (-1, 0).

S△abp ratio s△bpc=2 to 3.

Because s△abp and s△bpc are different bases and have the same height.

That is AP: PC = 2: 3.

oa？ +oc？ =ac？

ac=3√2

Oa=oc, so the angle oac is 45 degrees.

Then the distance from point P to Y axis =ac×3/5×cos angle oac=3√2×3/5×√2/2=9/5.

Distance from point P to X axis =ac×2/5×sin angle OAC = 3 √ 2/5 ×√ 2/2 = 6/5.

So the coordinates of point P are (-9/5, 6/5).

(2) According to the meaning of the question, set the parabolic analytical formula as y=ax? +bx+3

Replace (-3,0) (-2,0)

9a-3b+3=0

4a-2b+3=0

solve

a= 1/2，b=-5/2

y= 1/2x？ -5/2x+3

If there is a Q point, then the distance from Q point to Y axis = 1.

That is, when x= 1 or-1.

x=- 1，y=0

x= 1，y=5

Q (- 1, 0) or (1, 5)

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