Question

求解此题，

Answer 1

y'(t)=[6t(1+t^2)-6t^3]/(1+t^2)^2=6t/(1+t^2)^2,
x'(t)=[3(1+t^2)-6t^2]/(1+t^2)^2=3(1-t^2)/(1+t^2)^2,
y'(x)=y'(t)/x'(t)=2t/(1-t^2),
当t=2时，k=y'(x)=4/(-3)=-4/3. 排除CD。
又x(2)=6/5, y(2)=12/5. 设切线方程为4x+3y+b=0, 代入(6/5,12/5)得
24/5+36/5+b=0，解得b=-12.
因此选B. 法线方程检验一下就可以了。

Answer 2

x= 3t/(1+t^2)
dx/dt = 3[ (1+t^2) - 2t^2]/(1+t^2)^2 = 3(1-t^2)/(1+t^2)^2
y=3t^2/(1+t^2)
=3 -3/(1+t^2)
dy/dt = 6t/(1+t^2)^2
dy/dx
=(dy/dt)/(dx/dt)
=2t/(1-t^2)
dy/dx | t=2 = 4/(1-4) = -4/3
x(2) = 6/(1+4) = 6/5
y(2)= 12/(1+4) = 12/5
切线方程， t=2
y-y(2)= dy/dx | t=2 .[ x-x(2) ]
y-12/5 = -(4/3)(x- 6/5)
15y-36 = -4(5x-6)
20x+15y- 60=0
4x+3y -12 =0
法线方程 , t=2
y-y(2)= (3/4).[ x-x(2) ]
y-12/5= (3/4).( x-6/5)
5y-12= (3/4).( 5x-6)
20y-48= 3( 5x-6)
15x -20y +30=0
3x-4y +6 =0
ans : B