Fuzhou Ding Dian Recruitment

Solution: Solution: (1) From parabola y=ax2+bx+c through points A, B and C, we can get:

c=3a+b+c=0，9a+3b+c=0，

Solution: a = 1b =? 4，c=3，

The analytical formula of parabola is y = x2-4x+3.

(2) The passing point G is the GF⊥x axis, the vertical foot is F, and the coordinate of the point G is (m, m2-4m+3).

∫ point D(2,-1),

And ∵ b (3 3,0), c (0 0,3),

∴:CD = 25, BD=2, BC=32 from Pythagorean Theorem

∫CD2 = BC2+BD2，

△ CBD is a right triangle,

∴tan∠GAF=tan∠BCD= 13.

∫tan∠GAF = GFAF = 13

∴AF=3GF，

That is -3(m2-4m+3)=m- 1,

Solution: m 1= 1 (excluding), m2 = 83.

The coordinates of point G are (83, -59).

(3) The coordinate of point D is (2,-1),

△ Abd is an isosceles right triangle,

∴ Center E is the midpoint of line segment AB, that is, E (2 2,0) with radius of 1.

Let P(x 1, y1) (1< x1< 3, y 1≠0), M(3, y0) is the PF⊥x axis, and f is the vertical axis.

∵ Points A, P and M are on a straight line,

∴|y0||y 1|=2x 1？ 1, that is | y0 | = 2 | y 1 | x 1? 1.

∴tan∠meb=|y0|eb=2|y 1|x 1？ 1,

∫AB is the diameter,

∴∠APB=90，

∴∠PBA=∠APF，

∴tan∠PBA=tan∠APF=x 1？ 1|y 1|，

∴tan∠MEB？ tan∠PBA=2|y 1|x 1？ 1? x 1？ 1|y 1|=2。

Another solution: same as above, connect PE,

∫PE = 1，PF=|y 1|，EF=|x 1-2|，

In Rt△PEF, according to Pythagorean theorem, (x1-2) 2+y12 =1

That is,1-(x1-2) 2 = y12, …( 12 point),

∵tan∠PBA=|y 1|3？ X 1, …( 13 points)

∴tan∠MEB？ tan∠PBA=2y 12？ (x 12？ 4x 1+3)=2y 12 1？ (x 1？ 2)2=2.