(1)令x=0,则y=2, 所以,点C(0,2), ∵点M在直线y=-x+2上, ∴设点M的坐标为M(x,-x+2), 由勾股定理得CM==2, 整理得,x2=4, 解得x1=2,x2=-2, 当x1=2时,y1=-2+2=0, 当x2=-2,y2=-(-2)+2=4 ∴M(-2,4)或M(2,0), 当M(-2,4)时,设抛物线解析式为y=a(x+2)2+4, ∵抛物线过点C(0,2), ∴a(0+2)2+4=2, 解得a=-, ∴y=-x2-2x+2, 当M(2,0)时,设抛物线解析式为y=a(x-2)2, ∵抛物线过点C(0,2)点, ∴a(0-2)2=2, 解得a=, ∴y=x2-2x+2, ∴所求抛物线为:y=-x2-2x+2或y=x2-2x+2;
(2)∵抛物线与x轴有两个交点, ∴y=x2-2x+2不合题意,舍去. ∴抛物线应为:y=-x2-2x+2, 令y=0,则-x2-2x+2=0, 整理得,x2+4x-4=0, 解得x1=-2+2,x2=-2-2, ∵点A在B的左侧, ∴点A(-2-2,0),B(-2+2,0), ∴AB=(-2+2)-(-2-2)=4. |