∴k=-4，

∴ The analytic formula of hyperbola is y=-4/x,

∫ The distance from BC to X axis is four times that from point B to Y axis.

∴ Let the coordinate of point B be (m, -4m) (m > 0) and substitute it into hyperbolic analytical formula to get m= 1.

∴ parabola y = A(-2+bx+c (a < 0) passes through points A(-2,2), b (1 4), o (0 0,0).

∴4a-2b+c=2

a+b+c=-4

c=0，

Solution: a=- 1 b=-3 c=0,

So the analytical formula of parabola is y =-x2-3x;

(2) The analytical formula of parabola is y =-x 2-3x,

∴ Vertex E(-3/2, 9/4), the symmetry axis is x=-3/2,

∫B( 1，-4)，

∴-x^2-3x=-4，

Solution: x 1= 1, x2=-4,

∵C abscissa < 0,

∴C(-4,-4)，

∴S△ABC=5×6× 12= 15，

From the coordinates of A and B as (-2,2) and (1, 4), the analytical formula of straight line AB can be obtained as follows: y=-2x-2,

Let the symmetry axis of parabola intersect with AB at point F and connect BE, then the coordinate of point F is (-32, 1).

∴EF=9/4- 1=5/4，

∴s△abe=s△aef+s△bef= 1/2×5/4×3= 15/8；

(3)S△ABE= 15/8，

∴8S△ABE= 15，

∴ When point D coincides with point C, the condition is obviously satisfied;

When point D does not coincide with point C, the parallel CD passing through point C is AB, and its corresponding primary resolution function is y=-2x- 12.

Let-2x- 12 =-x 2-3x，

The solution is x 1=3, x2=-4 (truncation),

When x=3, y=- 18,

So there is still a point D(3,-18) that satisfies the condition.

To sum up, the coordinates of point D are (3,-18) or (-4, -4).