∵A, B is on a parabola, and the Y axis is the symmetry axis of the parabola.

∴ The abscissas of A and B are 2 and -2 respectively.

Substitute y= 1/4.

x？ + 1，a (2，2)，b (-2，2)，

∴M(0,2)，

(2) (1) If the intersection Q is QH⊥x axis and the vertical foot is H, then HQ=y and HP=x-t,

From △HQP∽△OMC, y/2=(x-4)/4 is obtained.

That is, t=x-2y,

Q(x, y) in y= 1/4.

x？ T =-1/2+1

x？

+x-2。

When point P coincides with point C, the trapezoid does not exist. At this time, t=-4, and x = 1 √ 5 is obtained.

When q coincides with b or a, the quadrilateral is a parallelogram, and X = 2.

The value range of ∴x is all real numbers of x ≠ 1 √ 5 and x ≠ 2;

(2) In two cases:

1) when cm > PQ, point p is on line segment oc.

∫CM∨PQ，CM=2PQ，

∴ The ordinate of point M is twice that of point Q, that is, 2=2( 1/4

x？ + 1)，x=0，

∴t=- 1/2

×0? +0-2=-2;

2) When cm < PQ, point P is on the extension line of OC.

∫CM∨PQ，CM=

PQ，

∴ The ordinate of point Q is twice that of point M, that is, 1/4.

x？ + 1=2×2,

Solution: x = 2 √ 3;

When x=-2√3, t=- 1/2(2√3)?

-2√3-2=-8-2√3,

When x=2√3

T = 2 √ 3-8。