(2) When the parabola moves down, the symmetry axis remains unchanged. Let the point where the axis of symmetry intersects the X axis be d, and let DB=n, then DA=n, OA=n- 1, and the coordinate of point A is 1-N, then the analytical formula of parabola is? =? -(x-n- 1)(x- 1-n)=-x & amp； sup2+2x+n & amp； sup2- 1,∴oc=n&； sup2- 1? ，DE=？ . n & ampsup2

∵? △DEB area = n &；; sup3/2,? The area of ladder code = (2n&; sup2- 1)/2,? Triangle COB area = (n&; sup3+n & amp； sup2-n- 1)/2

BCE area of triangle = (n&; sup2+n)/2？ , triangle ABC area = n (n&; sup2- 1)? , ∵ satisfies S△BCE=S△ABC In quadrilateral ABEC, ∴ column equation n (n&; sup2- 1)=(n & amp； Sup2+n)/2, n=3/2, then the coordinate of point B is (? 5/2,0), and the coordinates of point C are (0,0? 5/4)。 Then the analytical formula of straight line BC can be obtained as? y=- 1/2x+5/4

(3) Let the analytical formula of translation parabola be y=-(x-m)(x-n), then the coordinate of vertex e ((m+n)/2? ,? (m-n)& amp； Sup2/2), then the coordinate of C is (0, -mn)? The coordinates of point A are (m, 0) and the coordinates of point B are (n, 0).

△AOC area = m &；; sup2n/2

△EOC area =mn(m+n)/4

△OEB area = n (m-n)&; sup2/8

△COB area = m &；; sup2n/2

△CBE area =△EOC area? +? △OEB area? -? △COB area = n(n &；; Sup2- M&A; sup2)/8

Column equation n (n&; Sup2- M&A; sup2)/8=2? ×? m & ampsup2n/2

The solution is n = 3m. If n=-3m, then the vertex e? (-m, 4m & ampsup2), substituting y=-4x+3,

Get m 1=- 1/2, m2=3/2 (rejection) ∴n=3/2, ∴y=-(x+ 1/2)(x-3/2) or y =-. +x+3/4