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.

  • $\begingroup$ The cone is formed by all the lines passing through the origin and a point of the parabola. $\endgroup$ Jan 14, 2016 at 21:58
  • $\begingroup$ @Aretino But the cone should be "closed", while the parabola is open on one side, right? Even if this is true how will I know generate the equation of that cone? $\endgroup$
    – Stefan4024
    Jan 14, 2016 at 22:02
  • $\begingroup$ You might remember that the intersection of a cone and a plane can be a parabola. Its equation can be found in several ways, it all depends on what kind of knowledge of analytic geometry you have. $\endgroup$ Jan 14, 2016 at 22:06
  • $\begingroup$ How did you get $\bar y=\bar z$? $\endgroup$ Jan 15, 2016 at 10:09
  • $\begingroup$ Moreover: are you sure the text is correct? In particular: are you sure the coordinates of the cone vertex are $(3,-1,-2)$? I've plotted the parametric surface with GeoGebra and it doesn't look a cone. $\endgroup$ Jan 15, 2016 at 10:16

2 Answers 2



The equations defining the parabola can be rewritten as parametric equations as follows: $$ \cases{ x=t \cr \displaystyle y={1\over2}t^2-{1\over2}\cr \displaystyle z={1\over2}t^2+{1\over2}\cr } $$ The parametric equation of the line passing through the origin and a point $P(t)=(t,\ t^2/2-{1/2},\ t^2/2+{1/2})$ of the parabola is just $Q(s)=sP(t)$, that is: $$ \cases{ x=st \cr \displaystyle y={1\over2}s(t^2-1)\cr \displaystyle z={1\over2}s(t^2+1)\cr } $$ These are also the parametric equations of the cone, depending on two parameters $s$ and $t$. From these one finds $z+y=st^2$ and $z-y=s$, so that $z^2-y^2=s^2t^2=x^2$. The equation of the cone can be then expressed as $z^2=x^2+y^2$.

SHORT SOLUTION (valid only if the cone is a right circular cone).

The vertex of the parabola is $V=(0,-1/2,1/2)$ and this point must belong to a generatrix of the cone, which is then the line $x=0,\ z=-y$. The reflection of this generatrix around the axis of the cone must lie by symmetry on the $(y,z)$ plane and must be parallel to the plane $y-z+1=0$ (by definition of parabola as intersection of a plane and a cone). This second generatrix is then the line $x=0,\ z=y$.

It follows that the $z$ axis is also the axis of the cone, whose equation is thus $z^2=x^2+y^2$.

  • $\begingroup$ @Arentino Nice idea! Anyway I tried to solve another problem using this technique, but I came up against a problem. So can you help me a little bit with it? $\endgroup$
    – Stefan4024
    Jan 15, 2016 at 8:39


Two parabolic cylinders cut by a transverse plane.

The projections on $x,y,z $ planes are:

$$\begin{cases} x^2-2 z+1=0 \\ x^2 - 2 y -1 = 0 \\ y-z +1 =0 \end{cases} $$

The constant has to be set to zero in the last equation in order that it become the generator of a cone passing through origin:

$$ y=z $$


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