A and B are on a parabola, and the parabolic equation obtained by substituting the coordinates of A and B is as follows

y = x^2 - 2x - 3 = (x - 1)^2 -4

Then there are B =-2, C =-3, and the coordinates of point D are (1, -4).

② The slope of straight line AB is k 1 = 1.

To find a point F on the parabola that makes FA⊥AB, then the slope k2 of FA must satisfy k 1 k2 =-1 1, that is, k2 =-1.

Let F(m, n), then the slope of AF k2 = n/(m+ 1) =-1, that is, n = -m- 1.

Because point F is on a parabola, it exists.

n = (m- 1)^2 -4 = -m- 1

The solution is m = 2 or m =-1 (obviously omitted)

So there is a point f that satisfies the meaning of the question, and its coordinate is (2, -3).

③F is a moving point on the parabola below AB. In order to maximize the area of FAB, we need to find the farthest point from F to straight line AB.

Translate the straight line AB downward. The farther two straight lines are, the greater the distance is. After translation, the straight line A'B' and parabola have two intersections first, then another intersection (tangent), and finally separate. Obviously, when the straight line A'B' is tangent to the parabola, the downward translation is the largest, so the tangent point is the point f that satisfies the meaning of the question. Then set a straight line A'B' (the slope is 65438

There is only one intersection point between the straight line and the parabola y = x 2-2x-3, and the simultaneous equation can be solved at zero according to the discriminant to get t = 2 1/4.

The coordinates of point F are (3/2,-15/4).

④ In the early stage of the previous question, the coordinate of point F was (3/2,-15/4), and the AB equation was y = x+ 1.

Then the coordinates of point E are (1.5,2.5).

Because FE is perpendicular to the x axis,

If f is a right-angled vertex, then the parallel lines that pass through point f and make the X axis intersect the parabola intersect at another point p.

Substituting y =-15/4 into the parabolic equation, we can get x = 3/2 or 1/2, then the coordinate of point p is (1/2,-15/4).

If e is a right-angled vertex, then substituting y = 2.5 into the parabolic equation can get x = 1 plus or minus 6.5.