设直线L与C1相切于(x0,x0^2)
C1:y=x^2=>y'=2x=>L为y=2x0(x-x0)+x0^2=2x0x-x0^2
设直线L与C2相切于(x1,-(x1-2)^2)
C2:y=-(x-2)^2=>y'=-2(x-2)=4-2x=>L为y=(4-2x1)(x-x1)-(x1-2)^2=(4-2x1)x+x1^2-4
则有4-2x1=2x0,-x0^2=x1^2-4
解得x0=2,x1=0或x0=0,x1=2
所以L为y=4x-4或y=0