Find the equation of the cone whose vertex is at the origin and whose directing curve is given by the equations:
$$\begin{cases} x^2-2z+1=0 \\ y-z+1=0\end{cases} $$
We know that an eliptic cone is a surface of a revolution of a line around an axis. But I don't know how to use this, cause after all it seems that the equation of the directing curve defines a parabola in 3 dimensions. The parabola is obviously symmetric over the YZ-plane, but I don't know how to use it and whether it's helpful at all.
I tried to solve another problem using Arentino's solution, but I came up against problem. The question is similar except that the vertex is at $(3,-1,-2)$ and the directing curve is given by:
$$\begin{cases} x^2+y^2-z^2=1 \\ x-y+z=0\end{cases} $$
This time the intersection is hyperbola. Anyway by substituting $z=y-x$ in the first equation I got $2xy=1$, so the parametric equation of the hyperbola is given by:
$$\begin{cases} x=t \\ y=\frac 1{2t} \\ z = \frac{1-2t^2}{2t}\end{cases} $$
Now I change the coordinate system, such that I translate the original one, so the origin of the new one is the vertex of the cone, so:
$$\begin{cases} \bar{x}=x-3 \\ \bar{y}=y+1 \\ \bar{z}=z+2\end{cases} $$
So now I get for the cone:
$$\begin{cases} \bar{x}=st - 3s \\ \bar{y}=\frac {s}{2t} + s \\ \bar{z} = \frac{s-2st^2}{2t} + 2s \end{cases} $$
The only thing that I manage to get is that $\bar{y} = \bar{z}$ and using this that $\bar{x}=-2s$. But this forces $t=1$, which means that the intersection is a single point which isn't possible.