x=arctant=π/4
y=1+t^3=2
所以参数t=1
切线斜率
dy/dx
=(dy/dt)/(dx/dt)
=(arctant)'/(1+t^3)'
=(1/(1+t^2))/(1+3t^2)
=1/[(1+t^2)(1+3t^2)]
t=1时 dy/dx=1/8
点斜式
y-2=1/8(x-π/4)
8y-16=x-π/4
x-8y-π/4+16=0