设曲线有y=x+1/x-1在点(3,2)处的切线与直线ax+y+1=0垂直,则a等于?y=(x+1)/(x-1)=1+[2/(x-1)]y'=-2/(x-1)²当x=3时,y'=-2/(3-1)²=-1/2该点切线与直线ax+y+1=0垂直即它们的斜率乘积为-1-1/2×(-a)=-1解得a=-2,-1/2×(-a
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![设曲线有y=x+1/x-1在点(3,2)处的切线与直线ax+y+1=0垂直,则a等于?y=(x+1)/(x-1)=1+[2/(x-1)]y'=-2/(x-1)²当x=3时,y'=-2/(3-1)²=-1/2该点切线与直线ax+y+1=0垂直即它们的斜率乘积为-1-1/2×(-a)=-1解得a=-2,-1/2×(-a](/uploads/image/z/5352080-32-0.jpg?t=%E8%AE%BE%E6%9B%B2%E7%BA%BF%E6%9C%89y%3Dx%2B1%2Fx-1%E5%9C%A8%E7%82%B9%283%2C2%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E4%B8%8E%E7%9B%B4%E7%BA%BFax%2By%2B1%3D0%E5%9E%82%E7%9B%B4%2C%E5%88%99a%E7%AD%89%E4%BA%8E%3Fy%3D%28x%2B1%29%2F%28x-1%29%3D1%2B%5B2%2F%28x-1%29%5Dy%27%3D-2%2F%28x-1%29%26%23178%3B%E5%BD%93x%3D3%E6%97%B6%2Cy%27%3D-2%2F%283-1%29%26%23178%3B%3D-1%2F2%E8%AF%A5%E7%82%B9%E5%88%87%E7%BA%BF%E4%B8%8E%E7%9B%B4%E7%BA%BFax%2By%2B1%3D0%E5%9E%82%E7%9B%B4%E5%8D%B3%E5%AE%83%E4%BB%AC%E7%9A%84%E6%96%9C%E7%8E%87%E4%B9%98%E7%A7%AF%E4%B8%BA-1-1%2F2%C3%97%28-a%29%3D-1%E8%A7%A3%E5%BE%97a%3D-2%2C-1%2F2%C3%97%28-a)
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设曲线有y=x+1/x-1在点(3,2)处的切线与直线ax+y+1=0垂直,则a等于?y=(x+1)/(x-1)=1+[2/(x-1)]y'=-2/(x-1)²当x=3时,y'=-2/(3-1)²=-1/2该点切线与直线ax+y+1=0垂直即它们的斜率乘积为-1-1/2×(-a)=-1解得a=-2,-1/2×(-a
设曲线有y=x+1/x-1在点(3,2)处的切线与直线ax+y+1=0垂直,则a等于?
y=(x+1)/(x-1)=1+[2/(x-1)]
y'=-2/(x-1)²
当x=3时,y'=-2/(3-1)²=-1/2
该点切线与直线ax+y+1=0垂直
即它们的斜率乘积为-1
-1/2×(-a)=-1
解得a=-2,
-1/2×(-a)=-1
为什么是-a 这一步不太明白
设曲线有y=x+1/x-1在点(3,2)处的切线与直线ax+y+1=0垂直,则a等于?y=(x+1)/(x-1)=1+[2/(x-1)]y'=-2/(x-1)²当x=3时,y'=-2/(3-1)²=-1/2该点切线与直线ax+y+1=0垂直即它们的斜率乘积为-1-1/2×(-a)=-1解得a=-2,-1/2×(-a
直线ax+y+1=0可转化成斜截式:y=-ax-1,(-a为直线的斜率),所以是
-1/2×(-a)=-1,(斜率乘积为-1),
L1: y=k1x+b1 ,L2:y=k2x+b2
则L1⊥L2 的充要条件为:k1k2=-1
把一般式化为标准式Y=-aX-1,就ok了
已知:设曲线有y=x+1/x-1在点(3,2)处的切线与直线ax+y+1=0垂直,则a等于?
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