设直线为y=h,等边三角形边长为a,则高为h,且h=√3/2*a直线与抛物线相交,则有：h=2(x-2)^2=2x²-8x+8,即2x²-8x+8-h=0a=|x1-x2|=√[(x1+x2)²-4x1x2]=√[4²-4(4-h/2)]=√(2h)∴a²=2h=√3a => a=...