抛物线y=x²-2x-b与x轴分别交于A,B两点,求AB的距离
题目
抛物线y=x²-2x-b与x轴分别交于A,B两点,求AB的距离
求|AB| 的长
答案
设两交点横坐标分别为u,v
则u,v是方程x²-2x-b=0的两个根
|AB|=|u-v|=√(u-v)²=√[(u+v)²-4uv]=√(2²+4b)=2√(1+b)
举一反三
已知函数f(x)=x,g(x)=alnx,a∈R.若曲线y=f(x)与曲线y=g(x)相交,且在交点处有相同的切线,求a的值和该切线方程.
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