y=1⼀X在点(2,1⼀2)处的切线方程与法线方程

2024-12-03 14:42:44
推荐回答(1个)
回答1:

y=1/X在点(2,1/2)处的切线方程为x+4y-4=0
y=1/X在点(2,1/2)处的法线方程为4x-y-15/2=0
析:y的导数
=-1/x^2,则在点(2,1/2)处的切线斜率为-1/4
则y=1/X在点(2,1/2)处的切线方程为y-1/2=(-1/4)(x-2)
整理即得x+4y-4=0
切线与法线互相垂直,则在点(2,1/2)处的法线斜率为4
y=1/X在点(2,1/2)处的法线方程为y-1/2=4(x-2)
整理即得4x-y-15/2=0