∴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).