Question

求曲线x=4at/1+t^2,y=3at^2/1+t^2,t=2处的切线方程和法线方程

Answer 1

对t求导：
y'(t) = [(3at^2)'(1+t^2) - (3at^2)(1+t^2)']/(1+t^2)^2
= [6at(1+t^2) - 2t(3at^2)]/(1+t^2)^2
= [6at + 6at^3 - 6at^3]/(1+t^2)^2
= 6at/(1+t^2)^2
x'(t) = [(4at)'(1+t^2) - 4at(1+t^2)']/(1+t^2)^2
= [4a(1+t^2) - 8at^2]/(1+t^2)^2
= [4a + 4at^2 - 8at^2]/(1+t^2)^2
= [4a - 4at^2]/(1+t^2)^2
∴ dy/dx = 6at/(4a - 4at^2)
当t=2时,x = 8a/5,y = 12a/5
k = dy/dx = -1
∴切线方程为：y - 12a/5 = -(x - 8a/5) 即：x + y - 4a = 0
法线方程为：y - 12a/5 = x - 8a/5 即：x - y + 4a/5 = 0