Answer:
[tex]x=1+t,y=2,z=\frac{1}{3}+\frac{2}{3}t[/tex]
Step-by-step explanation:
[tex]z=\frac{x^2}{y^2-1}[/tex]
Point,P=(1,2,1/3)
Let x=t
y=2
[tex]z=\frac{t^2}{2^2-1}=\frac{t^2}{3}[/tex]
[tex]r(t)=<t,2,\frac{t^2}{3}>[/tex]
[tex]r'(t)=<1,0,\frac{2}{3}t>[/tex]
Substitute x=t=1
[tex]r(1)=<1,2,\frac{1}{3}>[/tex]
[tex]r'(4)=<1,0,\frac{2}{3}>[/tex]
The vector equation for the tangent line to C at the point P(1,2,1/3) is given by
[tex]r(t)=r(1)+tr'(1)[/tex]
[tex]r(t)=<1,2,\frac{1}{3}>+t<1,0,\frac{2}{3}>[/tex]
[tex]r(t)=<1+t,2,\frac{1}{3}+\frac{2}{3}t>[/tex]
Parametric equations of the tangent line to C at point P(1,2,1/3)
[tex]x=1+t,y=2,z=\frac{1}{3}+\frac{2}{3}t[/tex]