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 to cross the origin O, and get the straight line L. Let P be the moving point on the straight line L?

① Find the coordinates of point A; ?

(2) In a quadrilateral with points A, B, O and P as vertices,? There are rhombus, isosceles trapezoid and right-angled 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 coordinates of A are (-2, -4)?

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

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

Because the distance from point A to the X axis is 4?

So according to Pythagorean theorem: AB = OA = 2 √ 5?

Case 1: What 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 two sides of a diamond?

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: What 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 used as a waist of a trapezoid OB?

It is easy to find the analytical formula of the straight line L is y =-2x?

Since P is on L, let's assume that its coordinates are P(X, -2x)?

Because pa = ob = 4, the coordinates of point A are (-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: What 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?

Substitute OA = 2 √ 5 and AP4 = 8 √ 5/5 to get: OP4 = 6 √ 5/5?

So 5x 2 = 36/5 and x = 6/5 (-6/5 is truncated)?

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

P3(4/5, -8/5) can be found in the same way?

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

(3)?

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

At this time, a 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

Then 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.