若直线y=mx+1与曲线x2+4y2=1恰有一个交点,则m的值是__________.

若直线y=mx+1与曲线x2+4y2=1恰有一个交点,则m的值是__________.

答案

±

解析:将y=mx+1代入x2+4y2=1中,消去y得(1+4m2)x2+8mx+3=0,

Δ=64m2-12(1+4m2)=0.∴m=±.

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