As shown in the figure, the parabola y = x 2+4x intersects the X axis at point B and point O respectively, and its vertex is A, which connects AB. Move AB's straight line along the Y axis, and make it pass through the origin O, and get a straight line L. Let p be a moving point on a straight line L.

① Find the coordinates of point A;

② The quadrangles with points A, B, O and P as vertices include rhombus, isosceles trapezoid and right trapezoid. Please write the coordinates of the vertices p of these special quadrangles directly.

③ Let the area of a quadrilateral with points A, B, O and P as vertices be S, and the abscissa of point P be X, when 4+6 √ 2.

Solution:

(1) Because the parabolic equation is: y = x 2+4x.

Formula: y = (x+2) 2-4,

So the vertex coordinates of the parabola are (-2, -4). That is, the coordinate of A is (-2, -4).

(2) Let y = 0 and get X = 0 or -4.

So the coordinate of point B is (-4,0),

Because that distance from point a to the x axis is 4

So according to Pythagorean theorem, AB = OA = 2 √ 5.

Case 1: If the quadrilateral with vertices A, B, O and P is a diamond.

Because OB = 4 < 2 √ 5, OB < AB = OA.

So AB, OB or OA, OB can't be the edge of a diamond at the same time.

So it can only be OA and AB as the two sides of the diamond, and OB as the diagonal of the diamond.

So point p is the symmetrical point of point a about the x axis.

So the coordinate of point P is p1(-2,4).

(because oP 1/ab, p1must be on the straight line l).

Case 2: If the quadrilateral with vertices A, B, O and P is an isosceles trapezoid.

Because OB = 4 < 2 √ 5, OB < AB = OA.

So it can only be a waist of a trapezoid.

The analytical formula for finding the straight line L is y =-2x.

Since P is on L, it can be assumed that its coordinate is P(X, -2x).

Because PA = OB = 4, the coordinate of point A is (-2, -4).

So according to Pythagorean theorem:

(X+2)^2+(-2X+4)^2= 16

Solution: x = 2/5 or x = 2.

Because when x = 2, the quadrilateral ABOP is a parallelogram, which doesn't matter.

So x = 2/5.

At this time, the coordinate of P is P2(2/5, -4/5).

Case 3: If the quadrilateral with vertices A, B, O and P is a right trapezoid.

Obviously, it can only be AB as the bottom of the trapezoid.

The vertical lines passing through A and B are L and the vertical feet are P4 and P3 respectively (both coordinates can be expressed as (x, -2x)).

Let the distance from O to AB be h, and the following equation can be obtained according to the triangle area formula: AB * H = OB * 4.

So h = 8 √ 5/5, which means AP4 = AP3 = 8 √ 5/5.

According to Pythagorean Theorem: OA 2 = AP4 2+OP3 2

Substituting OA = 2 √ 5 and AP4 = 8 √ 5/5, we get: OP4 = 6 √ 5/5.

So it is 5x 2 = 36/5 and x = 6/5 (-6/5 rounded).

So the coordinate of point P is P4(6/5,-12/5).

Similarly, P3(4/5, -8/5) can be obtained.

That is, when the quadrilateral with vertices A, B, O and P is a right-angled trapezoid, the coordinate of point P is P3(4/5, -8/5) or P4(6/5,-12/5).

(3)

Because the analytical formula of the straight line L is: y =-2x, the coordinate dimension of p (x, -2x),

At this time, the quadrilateral with vertices A, B, O and P can be regarded as a trapezoid.

Upper bottom op = √ 5 | x |, lower bottom ab = 2 √ 5, and height h = 8 √ 5/5.

So s = (√ 5 | x |+2 √ 5) * (8 √ 5/5)/2.

=4|X|+8

So there are: 4+6 √ 2 < 4 | x |+8 < 6+8 √ 2.

So: 3 √ 2/2- 1

So when x > 0, the value range of x is 3 √ 2/2-1< x < 2 √ 2-1/2.

When x < 0, 3 √ 2/2- 1

So: x < 1-3 √ 2/2 or x > 1/2-2 √ 2.

Therefore, when x < 0, the value range of x is:1/2-2 √ 2 < x <1-3 √ 2/2.