已知,求在y=3(t^2)a/(1+t^2)x=3at/(1+t^2)处的切线方程和法线方程.

Question

已知 ,求在y=3(t^2)a/(1+t^2) x=3at/(1+t^2)处的切线方程和法线方程.

Answer 1

对t求导：
y'(t)=3a*[ 2t(1+t^2)-t^2*2t]/(1+t^2)^2=6at/(1+t^2)^2
x'(t)=3a*[1+t^2-t*2t]/(1+t^2)^2=3a(1-t^2)/(1+t^2)^2
因此y'(x)=y'(t)/x'(t)=2t/(1-t^2)
在点（x(t),y(t)）的切线方程即为：
y=y'*(x-x(t))+y(t)=2t/(1-t^2)* [x-3at/(1+t^2)]+3at^2/(1+t^2)
法线为：y=(t^2-1)/(2t)*[x-3at/(1+t^2)]+3at^2/(1+t^2)