3

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.

user161997
  • 99
  • The $y$ should have its $x$ replaced by $t$ in the parametrization. Looks like $C$ is itself a parabola lying in the plane $z=1-x$ to me. [the only other alternatives, an ellipse or two crossed lines, can be ruled out since $y$ is not bounded, and it is also not two intersecting lines.] – coffeemath Jul 07 '14 at 00:51
  • Do you need more details about the parabola? Or is it enough to just know whether it is one or not? – coffeemath Jul 07 '14 at 00:57
  • @coffeemath I see what you're saying I think, although I don't know how to show that it's not two intersecting lines. Also, it could also be a hyperbola too right? – user161997 Jul 07 '14 at 01:02
  • Yes, could be a hyperbola. (Forgot that one). It should be relatively easy to rule out all but parabolas. – coffeemath Jul 07 '14 at 01:04

2 Answers2

2

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.

Lubin
  • 62,818
  • 2
    Maybe you could answer a slightly related question. Would it be possible to have $C$ such that the projection onto the xy plane was a parabola and $C$ itself was not a parabola? Is the projection onto the xy plane sufficient to show that $C$ is a parabola? – user161997 Jul 07 '14 at 02:44
  • The question is whether a nonparabola can project orthogonally onto a parabola. Certainly an ellipse (or circle) can’t, because it is compact, and its image must be compact. So the question is whether a hyperbola can project to a parabola. Consider the minor axis, that’s the axis of symmetry between the two branches. It will project to a line, and the two branches of the hyperbola will project to opposite sides of this line (unless the whole figure is projecting to a single line, which we would want to exclude). So the projection of a hyperbola will be disconnected, and thus not a parabola. – Lubin Jul 07 '14 at 16:00
1

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.

Tom Oldfield
  • 13,034
  • 1
  • 39
  • 77
  • Can you say what $\delta_{i,j}$ is? I'm not familiar with that notation. Also, is $\langle e_i,e_j \rangle$ the inner product? – user161997 Jul 07 '14 at 01:06
  • That works, but the plane has many orthogonal bases. Can you explain how you picked the right one? – David Jul 07 '14 at 01:07
  • @user161997 It's the Kronecker delta: http://en.wikipedia.org/wiki/Kronecker_delta

    What I'm trying to say, is that $e_1$ and $e_2$ give an orthonormal basis for the plane, in the same way we use $(1,0)$ and $(0,1)$ as a basis for the $x$ and $y$ vectors in the usual Euclidean plane, which can be considered as the plane $z = 0$ in $\mathbb{R}^3$.

     – Tom Oldfield Jul 07 '14 at 01:09
  • @David The point is, as long as the basis is orthonormal, it doesn't matter which we choose. Conic sections will all be of the same form with respect to any orthonormal basis, since changing between them is just an isometry, and isometries map circles to circles, parabolas to parabolas, etc. You can prove this from the algebraic definition of each conic section, or whichever geometric one you prefer. e.g. a circle is the set of points a given distance from some origin and this is co-ordinate independent. – Tom Oldfield Jul 07 '14 at 01:11
  • @Tom sure, but if the parabola is not aligned with the coordinate axes then its equation will be harder to recognise, it won't just be $u=a+bv^2$. – David Jul 07 '14 at 01:12
  • @David Quite possibly. If you're asking how I picked a "nice" basis then the answer I guessed and it worked out fine. Even if my choice of basis wasn't nice, all we really need to know about the resulting equation $ay^2+bxy+cy^2+dx+ey+f=0$ is which coefficients are $0$, which should always be obvious. – Tom Oldfield Jul 07 '14 at 01:17
  • @David Actually, once I started, I realised $(0,1,0)$ was a direction in the plane and this made it a lot easier. In general picking a nice basis might be harder, but I imagine it should be possible to have a good try from inspection. – Tom Oldfield Jul 07 '14 at 01:20
  • Actually, sorry, we need to know the discriminant of the algebraic form, not which coefficients are $0$. But the point still stands. – Tom Oldfield Jul 07 '14 at 01:28