(2) Therefore, it is easy to find the coordinate of point B as (3,3 √ 3).
And by parabola y=ax2-2√? 3x? =x(ax-2√? 3) = 0,x 1=0,x2=2√? 3/a=2(x 1 should be the abscissa of a)
So a=√? three
? So the parabola is y=√? 3x^2-2√? 3x? Therefore, the generation check of b coordinate satisfies that b is on a parabola.
(3) Can calculate the coordinates of point D (1,-√? 3)
tan∠OAB=-tan∠BAx=-3√? 3=-tan∠OPD=tan∠APD
? Let P(x, 0), then tan∠OPD =∞? 3/(x- 1)=3√? 3, so x=4/3.
(4) As shown in the figure below, because DE = 1 and AP = 1, P( 1, 0) is obtained.