Find the equation of the cylinder that has directrix the curve: $x(t)=t, y(t)=t^2/2, z(t)=0$ and the generatrix is parallel to the line $${x-1\over 1}={y+2\over 1}={z\over 3}$$
I would really appreciate if you can help me how to solve this problem; are there any books of analytic geometry or links that have this kind of problems about cylinders?
A direction vector of the straight line given is $(1,1,3)$. We are on the surface if $(x,y,z)+(u,u,3u)=(t,\frac{t^2}{2},0)$ for some $t$ and $u$. Intuitively: sitting on the surface (at $(x,y,z)$) and stretching a vector (parallel to, say $(1,1,3)$) towards the $xy$ plane, the vector has to reach it at a point for which the directrix equation holds:
We have, therefore, the following system of equations:
$$ \begin{equation} x+u=t\\ y+u=\frac{t^2}{2}\\ z+3u=0 \end{equation} .$$
From the first two equations we get that
$$t^2-2t+2(x-y)=0$$
therefore $$t_{1,2}=1\pm\sqrt{1-2(x-y)}.$$
($y\ge x-\frac{1}{2}$, since $t_{1,2}$ have to be real.) Then, because of the first equation,
$$u=1\pm \sqrt{1-2(x-y)}-x.$$
So
$$z(x,y)=-3\left(1\pm \sqrt{1-2(x-y)}-x.\right) \text{ if }\ y\ge x-\frac{1}{2}.$$ (If one checks the $\pm$ signs above, it turns out that only the negative sign works.)
I hope that this solution is OK. Unfortunately I cannot point at any general reference.