It is known that the parabola y = x2 + (2n-1) x + n2-1 (n is a constant). (1) when the parabola passes through the coordinate origin and the vertex is in the fourth quadrant, the corresponding functional relationship is obtained; (2) let a be a moving point on the parabola determined by (1) below the X axis and to the left of the symmetry axis, pass through a as a parallel line of the X axis, intersect the parabola at another point D, and then make ab ⊥ X axis at B, DC ⊥ X axis at C. ① when BC = 1, find the perimeter of rectangular ABCD; ② ask whether there is a maximum perimeter of rectangular ABCD? If it exists, request the maximum value and point out the coordinate of point A. if it does not exist, please explain the reason

It is known that the parabola y = x2 + (2n-1) x + n2-1 (n is a constant). (1) when the parabola passes through the coordinate origin and the vertex is in the fourth quadrant, the corresponding functional relationship is obtained; (2) let a be a moving point on the parabola determined by (1) below the X axis and to the left of the symmetry axis, pass through a as a parallel line of the X axis, intersect the parabola at another point D, and then make ab ⊥ X axis at B, DC ⊥ X axis at C. ① when BC = 1, find the perimeter of rectangular ABCD; ② ask whether there is a maximum perimeter of rectangular ABCD? If it exists, request the maximum value and point out the coordinate of point A. if it does not exist, please explain the reason

(1) When n = 1, y = x2 + X is obtained, and the vertex of the parabola is not in the fourth quadrant. When n = - 1, y = x2-3x is obtained, and the vertex of the parabola is in the fourth quadrant. The functional relationship is y = x2-3x; (2) by y = x2-3x, let y = 0, then x2-3x = 0, and the solution is X1 = 0, X2 = 3. The other intersection of the parabola and the X axis is (3, 0) Its vertex is (32, − 94), its axis of symmetry is a straight line x = 32, and its approximate position is shown in the figure. ① ∵ BC = 1, it is easy to know that OB = 12 × (3-1) = 1. The abscissa of point a is x = 1, and point a is on the parabola y = x2-3x, and the ordinate of point a is y = 12-3 × 1 = - 2.. AB = | y | = | - 2 | = 2. The circumference of rectangular ABCD is 2 (AB + BC) = 2 × (2 + 1) = 6. ② ∵ point a is on the parabola y = x2-3x, So let a be (x, x2-3x), B be (x, 0). (0 ＜ x ＜ 32) ‖ BC = 3-2x, a be under the X axis, ‖ x2-3x ＜ 0, ‖ AB = | x2-3x | = 3x-x2 ‖ the perimeter of rectangle ABCD, C = 2 [(3x-x2) + (3-2x)] = - 2 (X-12) 2 + 132, ∵ a = - 2 ＜ 0, the opening of parabola is downward, the quadratic function has the maximum value, ‖ when x = 12, the maximum perimeter C of rectangle ABCD is 132 The coordinate of point a is a (12, − 54)

Given the parabola y = x & sup2; + (2n-1) x + n & sup2; - 1 (n is a constant term), when the parabola passes through the coordinate origin and the vertex is in the fourth quadrant, find the function
(2) A is a moving point below the x-axis of the parabola and on the left side of the symmetry axis, passing through a to make a parallel line of the x-axis, intersecting the parabola at another point D, and then making ab ⊥ X-axis at B, CD ⊥ X-axis at C. ① when BC = 1, find the perimeter of rectangular ABCD. ② ask whether the perimeter of rectangular ABCD has the maximum value? If it exists, request the maximum value, and point out the coordinates of point a at this time. If it does not exist, explain the reason.

Because the parabola passes through the coordinate origin
So substituting (0, 0) into (0, 0) gives:
n² - 1 = 0
N = 1 or n = - 1
When n = 1, y = x & # 178; + X vertex in the third quadrant, rounding off
When n = - 1, y = x & # 178; - 3x vertex is in the fourth quadrant
So the parabola analytic formula y = x & # 178; - 3x

It is known that the parabola y = x2 + (2n-1) x + n2-1 (n is a constant). When the parabola passes through the origin and the vertex is in the fourth quadrant, the corresponding functional relation is obtained

From the known conditions, we get n2-1 = 0, and solve this equation, we get N1 = 1, N2 = - 1. When n = 1, we get y = x2 + X, and the vertex of this parabola is not in the fourth quadrant; when n = - 1, we get y = x2-3x, and the vertex of this parabola is in the fourth quadrant. Therefore, the functional relationship is y = x2-3x

It is known that the line y = x intersects with the parabola y = ax & # 178 at the origin O and point A. after moving the line y = x three units horizontally, it intersects with the parabola y = ax & # 178 at B, C,
If Ao: BD = 1:2, find the value of A

When a > 0, the line y = x shifts 3 units to the left and intersects the x-axis at point D
The line y = x-3 and parabola y = ax & # 178 intersect with B and C
So the coordinate of point D is (- 3,0)
Because the line y = x and the parabola y = ax & # 178 intersect with point a
So the coordinates of a are (1 / A, 1 / a)
Make a vertical line of X axis through two points a and B respectively, and the vertical foot is e and F
Then AE = 1 / A,
In RT △ BDF and RT △ AOE,
Because BD ‖ OA
Therefore, BDF = AOE
∠BFD=∠AEO=Rt∠
So RT △ BDF ∽ RT △ aoe
So BD ∶ OA = BF ∶ AE
Because Ao: BD = 1:2
So BF ∶ AE = 2 ∶ 1
So the ordinate of point B is 2 / A
Because point B is on the line y = x-3,
So the coordinates of B are (2 / a-3,2 / a)
Substituting the coordinates of point B into the parabola y = ax & # 178;, we get
2/a=a(4/a^2－12/a+9)
It is reduced to 9A ^ 2-12a + 2 = 0
So a = (2 ± √ 2) / 3
When a