切线方程与法线方程

Question

如题所述

Answer 1

x=2e^t
dx/dt = 2e^t

y=e^(-t)
dy/dt =-e^(-t)

dy/dx = dy/dt /dx/dt =- (1/2)e^(-2t)
dy/dx|(t=0) = -1/2

t=0
x=2e^t = 2
y=e^(-t) = 1

切线方程 t=0
y-1 = -(1/2)(x-2)

法线方程 t=0
y-1 = 2(x-2)