Determine whether the intersection of surfaces is a parabola 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. 

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.
