∴ k 2+k = 0, k = 0 (irrelevant, omitted), k=- 1.

The analytical formula of parabola is: y =-x 2+2 √ 3x =-(x-3) 2+3.

Vertex b (√ 3,3)

(2) Easy to obtain: A (2 √ 3,0), the symmetrical point A' (-2 √ 3,0) about the Y axis, connecting A'B and the Y axis at P.

Let the symmetry axis of parabola intersect the X axis at m, then A'M=3√3 and BM=3.

∴tan∠ba'm=3/(3√3)=√3/3,∴∠ba'm=30

∴OP=OA'*tan30 =2,∴P(0,2)

(3)∫AC∨BP, ∴ OAC = ∠ Ba 'o = 30, ∴OC=2, that is, C(0, -2).

∴AP=AC, so the internal q required by Δ δACP is on the X axis and on the angular bisector of ∞∠PCA.

∴∠ qco = 30, OQ = oc * tan30 = 2 √ 3/3, which is the required point q (2 √ 3/3,0).