问题 填空题

若抛物线f(x)=x2+ax与直线f'(x)-1-y=0相切,则此切线方程为 ______.

答案

∵f(x)=x2+ax

∴f'(x)=2x+a

则f'(x)-1-y=0即2x-y+a-1=0

∵抛物线f(x)=x2+ax与直线f'(x)-1-y=0相切

y=x2+ax
y=2x+a-1
即x2+(a-2)x+1-a=0只有一解

即△=(a-2)2-4(1-a)=0

解得a=0

∴此切线方程为2x-y-1=0

故答案为:2x-y-1=0

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