(2)① According to the properties of similar triangles, the abscissa expression of point E is the abscissa expression of point G, and the expression of ordinate is obtained by substituting it into the quadratic resolution function, thus transforming the maximum problem of line segment into the solution of the maximum problem of quadratic function.
② If an isosceles triangle is formed and two of the three sides are equal, it can be discussed in three situations: EQ=EC, EC=CQ and EQ=EC. If the time of the two cases is the same, then the length of the three sides is the same, which is an isosceles triangle.
Solution:
(1) Because the abscissa of point B is 4, the ordinate of point D is 8, and the coordinate of point A is (4,8). ( 1).
Substitute the coordinates of a (4 4,8) and c (8 8,0) into {16a+4b=8 64a+8b=0 of y=ax2+bx to get a=- 1/2 and b=4.
The analytical formula of parabola is: y =-1/2x2+4x; (3 points)
(2)① In Rt△APE and Rt△ABC, tan∠PAE= PE/AP= BC/AB, that is, PE/AP = 4/8.
∴PE= 1/2AP= 1/2t。 PB=8-t。
The coordinate of point e is (4+ 1/2t, 8-t).
∴ The ordinate of point G is:-1/2 (4+1/2t) 2+4 (4+1/2t) =-1/8t2+8. (5 points)
∴eg=- 1/8t^2+8-(8-t)=- 1/8t^2+t.
∫- 1/8 < 0, ∴ When t=4, the longest line segment eg is 2. (7 points)
(2) * * * there are three moments. (8 points)
(1) when EQ=QC,
Because Q(8, t), E(4+ 1/2t, 8-t), QC=t,
So according to the formula of the distance between two points, we can get:
( 1/2t-4)^2+(8-2t)^2=t^2.
Arrange to 13t 2- 144t+320 = 0,
The solution method is t= 40/ 13 or t= 104/ 13=8 (at this time, E and C coincide and cannot form a triangle, so it is discarded).
(2) When EC=CQ,
Because E(4+ 1/2t, 8-t), c (8 8,0), QC=t,
So according to the formula of the distance between two points, we can get:
(4+ 1/2t-8)^2+(8-t)^2=t^2.
T 2-80t+320 = 0, t=40- 16 radical number 5, t=40+ 16 radical number 5 > 8 (at this time, q is not on the edge of the rectangle, so it is discarded).
(3) when EQ=EC,
Because Q(8, t), E(4+ 1/2t, 8-t), C(8, 0),
So according to the distance formula between two points, (1/2t-4) 2+(8-2t) 2 = (4+1/2t-8) 2+(8-t) 2,
The solution is t=0 (at this time, Q and C coincide and cannot form a triangle, so it is discarded) or t = 163.
So t 1= 16/3, t2= 40/ 13, t3=40- 16, and the number of roots is 5. ( 1 1)
Comments: The solution of parabola is an analytic function. Usually, the undetermined coefficient method is used, that is, the equations are listed first, and then the unknown coefficients are obtained. This method is more suitable for this topic. For the moving point problem and extreme value problem in the finale, first, according to the conditions, use "static braking" to express their respective coordinates. If a quadratic function can be formed, the extreme value can be obtained by formula or vertex coordinate formula.