Let $C$ be the curve of intersection between the cone $z=\sqrt{x^2 + y^2}$ and the plane $z=1-x$. Is $C$ a parabola?
I can see that letting $x=t$ we have parametric equations
$x=t\\y=\pm\sqrt{1-2t}\\z=1-t$
and so the projection onto the $xy$ plane is a parabola. But I don't know how to see whether $C$ itself is a parabola.
On
If you want to go back to the classical definition of conic sections, all of them are what you get by intersecting a right regular (circular) cone with a plane that doesn’t contain the vertex of the cone. The generatrices of the cone are the straight lines lying in the cone, all passing throught the vertex. Since it’s a right cone, the generatrices all make the same acute angle $\theta$ with the axis. In your example, $\theta=45^\circ$.
Now, in general, if your plane is perpendicular to the axis, you get a circle; if the plane makes an angle to the axis that’s strictly between $90^\circ$ and $\theta$, you get an ellipse; if less than t$\theta$, you get a hyperbola. But if the plane is parallel to one of the generatrices, i.e. if the angle between the plane and the axis is $\theta$, you get a parabola.
In your case, you have a generatrix given by the equations $\{z=-x, y=0\}$, and your plane is certainly parallel to this line. So your intersection is a parabola.
Note that,as written, $z = \sqrt{x^2+y^2}$ is only a half cone, you need $z^2=x^2+y^2$ for the full thing (or an unwieldy $\pm$).
Given this, $\mathcal{C}$ is the solution set of $z = 1-x$ and $y^2 = 1-2x$. This is contained in the plane $x+z=1$ and we want to express this curve in terms of orthogonal co-ordinates for this plane. We do this by noting that any point on the plane can be written as $(1,0,0)+ue_1+ve_2$ where $e_1 = \frac{1}{\sqrt{2}}(1,0,-1)$ and $e_2 = (0,1,0)$, and $\langle e_i,e_j\rangle = \delta_{i,j}$. Then $v = y$ and $ x = 1+\frac{u}{\sqrt{2}}$. Thus $\mathcal{C}$ is the solution to $v^2=-1-\sqrt{2}u$ which is indeed a parabola.