According to the nature of folding, BC=CD, angle B= angle D.
So CD=OA, angle D= angle AOC, angle CED= angle AEO.
So the triangle CED is equal to the triangle AEO.
So AE=CE, and because C (-8,0 0) A (0 0,4), OE+CE=OE+AE=8, OA=4.
Because in the right triangle AEO, AE 2 = OE 2+OA 2.
Therefore, (8-OE) 2 = 4 2+OE 2. If we solve this equation, we get OE=3.
So e (-3,0)
(2) If D is passed, make DF perpendicular to X axis and F (draw by yourself).
Then cos angle FED= cos angle OEA=OE/EA=3/5 because angle FED= angle OEA.
So in the right triangle DFE, EF=DE*cos angle FED=3*3/5=9/5.
DF= root number (de 2-ef 2) = 12/5, OF=OE+EF=24/5.
So D(-24/5,-12/5)
Then let the analytical formula of parabola passing through O, D and C be y = AX 2+BX+C.
By solving the equations, the analytical formula of parabola is y = 5/32x 2+5/4x.
(3) Because y = 5/32x 2+5/4x = 5/32 (x+4) 2-5/2.
So the vertex f is (-4, -5/2)
The following points were discussed:
(i) When t is less than 4, that is, P is on the OA segment, it is easy to know that PF does not intersect with any line segment of triangular FAC, so there is no PF to divide the area of triangular FAC into two parts: 1: 3.
(2) When t is greater than or equal to 4 and less than 12, that is, P is on the AB segment, and the straight line PF passes through points AC and F. At this time, if PF wants to divide the area of triangular FAC into two parts: 1: 3, it only needs PF to divide the line segment AC into two parts: 1: 3. Let PF and AC intersect with h, and the intersection with h makes HI perpendicular to the x axis at I (.
Let the straight line AC be y=kX+h, and bring it into A and C. It is easy to find out that AC is y= 1/2x+4. From the coordinates of A and C, it is easy to find the root number 5 of CA=4, which will be discussed in different situations:
Case 1: when ch: ah = 1: 3, ch: (ch+ah) = 1: 4, that is, CH:CA= 1:4. At this time, CH=CA/4= radical number 5, then hi = (ch *.
X=-6, so H (-6, 1), let FH be y=K 1x+B 1, and bring in the coordinates of f and h, it is easy to find that FH is y=- 1.75x-9.5, and y=4.
So t 1=82/7.
Case 2: when CH: AH = 3: 1 and CH:(CH+AH)=3:4, we can get h (-2,3) by the above method, let FH be y=K2x+B2, and bring in f and h, we can get y=2.75x+8.5. So p (-18/1,4), so AP =18/1,OA=4, so T2 = 62/1/.
(3) When t is greater than or equal to 12 and less than 16, that is, when P is in BC section, HC is always less than 1/4 times AC, so it is easy to know that there is no t to divide the area of △ △FAC into 1: 3.
To sum up, there are two eligible t***, namely 82/7, 62/ 1 1.
Because it is difficult to input mathematical symbols, many places are described in language, and the calculation is in a hurry, which may make mistakes. I hope LZ can check for itself.