It is known that the parabola y = x & # 178; + (2n-1) x + n & # 178; - 1 (n is a constant term) It is known that the parabola y = x & # 178; + (2n-1) x + n & # 178; - 1 (n is a constant term) (1) when the parabola passes through the origin and the vertex is in the fourth quadrant, the corresponding functional relationship is obtained. (2) Suppose that point a is a moving point on the parabola determined by (1), and it is located below the x-axis and on the left side of the symmetric state, passing through point a as a parallel line of the x-axis, intersecting the parabola with another point D, making ab ⊥ x-axis and point B, DC ⊥ X-axis at point C, then (1) when BC = 1, find the perimeter of rectangular ABCD (2) whether there is a maximum perimeter of rectangular ABCD? If it exists, calculate the maximum value and the coordinate of point A. if it does not exist, please explain the reason,

It is known that the parabola y = x & # 178; + (2n-1) x + n & # 178; - 1 (n is a constant term) It is known that the parabola y = x & # 178; + (2n-1) x + n & # 178; - 1 (n is a constant term) (1) when the parabola passes through the origin and the vertex is in the fourth quadrant, the corresponding functional relationship is obtained. (2) Suppose that point a is a moving point on the parabola determined by (1), and it is located below the x-axis and on the left side of the symmetric state, passing through point a as a parallel line of the x-axis, intersecting the parabola with another point D, making ab ⊥ x-axis and point B, DC ⊥ X-axis at point C, then (1) when BC = 1, find the perimeter of rectangular ABCD (2) whether there is a maximum perimeter of rectangular ABCD? If it exists, calculate the maximum value and the coordinate of point A. if it does not exist, please explain the reason,

(1)n^2-1=0
n^2-1
N = 1 or n = - 1
When the vertex is in the fourth quadrant
So n = - 1
y=x²-3x
(2) The symmetry axis of y = x & # 178; - 3x is x = 3 / 2
When BC = 1
Then ad = 1
The coordinate of a is (1, ya)
The coordinate of D is (2, YD)
ya=1^2-3*1=-2
yd=2^2-3*2=-2
The coordinates of a are (1, - 2)
The coordinates of D are (2, - 2)
Perimeter = (2 + 1) * 2
=6
(2) Let the coordinates of a be (x, X & # 178; - 3x) with larger perimeter
Perimeter L = 2 * [2 * (3 / 2-x) - X & # 178; + 3x] has maximum solution
L=2*[3-2x -x²+3x ]
=2*[3-x²+x ]
=-2(x²-x -3)
=-2(x-0.5)²+6.5
The coordinates of a are (0.5, - 1.25)
The maximum solution is 6.5