已知曲线y=x2求过点B(2,3)且与曲线相切的直线方程

2025-03-23 11:33:46
推荐回答(1个)
回答1:

设切点为(t,t^2)
y'=2x, y'(t)=2t
切线为y=2t(x-t)+t^2=2tx-t^2
代入B(2,3)到切线: 4t-t^2=3
(t-1)(t-3)=0
t=1, 3
当t=1时,切线为y=2x-1
当t=3时,切线为y=6x-9