∵y=2x+1,∴x=(y-1)/2
将x=(y-1)/2代入y²=12x中,得：y²=6(y-1)
即：y²-6y+6=0
设A(x1, y1),B(x2, y2),则y1+y2=6, y1y2=6
∴(y1-y2)²=(y1+y2)²-4y1y2=12
而x1=(y1-1)/2,x2=(y2-1)/2
∴x1-x2=(y1-y2)/2
∴(x1-x2)²=(y1-y2)²/4=3
∴|AB|=√[(x1-x2)²+(y1-y2)²]=√15
即：|AB|=√15
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