∵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。