Let the coordinate of E be (m, 0), because the slopes of AE and CE are the same, there is (2px1-0)/(x1-m) = (-2px2-0)/(x2-m).

Simplification, m=2x 1x2/(x 1+x2)

Similarly, the coordinate of d can be set to (n, 0), because the slopes of AD and BD are the same, with (2px1-0)/(x1-n) = (2px2-0)/(x2-m).

Simplify and get n= -2x 1x2/(x 1+x2).

Since the absolute values of m and n are the same and the signs are different, O is the midpoint.

Question 2: Let the coordinate of C be (x, y).

Since the directrix is x= -2 and the fixed point is the midpoint between the focus and the directrix, we can get that the focus coordinate is (2x+2, y).

The parabola always passes through the fixed point A (2,0), so the distance from the point A (2,0) to the focus is equal to the distance from the directrix.

We get: (2x+2-2) 2+y 2 = [2-(-2)] 2, which is simplified as x 2/4+y 2/16 =1,and it is an ellipse with the focus on the y axis.

It doesn't exist. This is used for painting. When the opening angle of any two points on the ellipse to B is maximum, the two points on the X axis, namely (2,0) and (-2,0), are still acute angles, so they do not exist.

Thirdly, it is obvious that these two straight lines can't be the X axis and the Y axis, because they have only 1 intersection with the parabola.

Let a straight line be y=kx and another straight line be y =-x/k.

Simultaneous equations y=kx y=-x/k respectively.

y^2 =6x y^2=6x

Solve the other two points (6/k 2, 6/k), (6/k 2, -6k).

Their midpoint is (3/k 2+3k 2, 3/k-3k) = (x, y).

Then, x and y are transformed into each other and k is eliminated.

y^2 =9( 1/k^2+k^2-2)= 3x- 18

That is: y 2 y 2+18 = 3x This is a parabola.

Answer finished, I hope you are satisfied.