Consider the surface $S$ given by $x^3+ zy^3=8z^2$, let $C$ be an intersection curve of the surface $S$ and the cylinder $x^2+y^2=4$, find a set of parametric equations for the tangent line to the curve $C$ at $(0,2,1)$
Note that the equation of the tangent plane at $(0,2,1)$ on the surface $S$ is given by $3y-2z=4$
My usual method was to substitute the equations into each other and hopefully terms will easily cancel each other out which then I can parameterise the intersection curve $C$ easily, but looks like this cannot be done easily.
Instead, I tried to use the tangent plane to find the direction vector, in this case, I can find the normal vector on $(0,2,1)$ on the cylinder using the del operator. I obtained $<0,4,0>$, crossing this with the normal vector $<0,3,-2>$ yields me with the well needed direction vector $<-8,0,0>$.
I got the answer while just randomly guessing around and using what I could use, I tried to plot the graphs out using Geogebra but I am still unable to understand why this works.
I will really appreciate it if someone can explain why I can do this to me to help me visualise this better! Many thanks!