I was successful in deriving the equation for an elliptical double-napped cone in rectangular coordinates. All I did was define a line with slope $a$ on the xy-plane, and another line of slope $b$ on the xz-plane. Then, I used values of both functions concurrently to trace an ellipse (with parameters obtained from the linear functions) on all planes parallel to the yz-plane. With that, I was able to obtain the equation I was looking for. However, how can I prove that the conditions I used were enough to define the surface I was looking for? i.e. how do I know that there isn't another surface that $1)$ makes traces of ellipses to all planes parallel to the yz-plane and $2)$ makes a trace of two intersecting lines with the xy and xz planes and isn't a cone?

I'm pretty sure that they are sufficient conditions, since I derived the correct equation, but say I didn't have/know the equation? Thank you.


1 Answer 1


I don't think you have enough information. By specifying that the intersection of the surface with the xy- and xz- planes are two lines, this fixes four distinct points for each plane $P$ that is parallel to yz- plane (except the plane through the vertex of the cone).

In other words, for a fixed $x_0$, you know that the points $(x_0,\pm ax_0,0)$ and $(x_0,0,\pm bx_0)$ (assuming you take the two lines with slope $\pm a$ in the xy-plane and the lines with slope $\pm b$ in the xz-plane; you can easily modify this to fit four different slopes.

Now, you know that the figure is an ellipse on the plane $P = \{ x = x_0\}$. So we have that on that plane, using the yz-coordinates, you have that the general equation for an ellipse is

$$ \alpha y^2 + \beta z^2 + \gamma yz + \delta y + \epsilon z + 1 = 0 $$

with the requirement that $\gamma^2 < 4\alpha \beta $ and $\alpha,\beta \neq 0$. In other words, the ellipse has five degrees of freedom: the coordinates of the two foci (2 parameters each, so four total), and the eccentricity.

Since you only specified 4 points, there is one remaining free parameter. In terms of the above by specifying that four points, you showed that $\alpha = -(ax_0)^{-2}$, $\beta = -(bx_0)^{-2}$, $\delta = \epsilon = 0$. So if you let $\gamma = \gamma(x) = \frac{1}{x^2}\tilde{\gamma}(x)$ be any function of $x$ satisfying the requirement that

$$ \tilde{\gamma}(x)^2 < \frac{4}{a^2 b^2} $$

So we can define a surface $S$ by the implicit function

$$ \frac{y^2}{a^2} + \frac{z^2}{b^2} + \tilde{\gamma}(x) yz = x^2 $$

(in particular, we can choose $\tilde{\gamma}(x) = \frac{1}{ab}\sin x$). This will give a surface satisfying the properties that

  1. Its intersection with any plane that is parallel to the yz-plane is an ellipse
  2. Its intersection with xz-plane is two lines with slope $\pm b$
  3. Its intersection with the xy-plane is two lines with the slope $\pm a$.
  4. It is not a cone. (The implicit function will give you a cone if and only if $\tilde{\gamma}(x)$ is a constant function.)

Remark: the construction above is achieved by rotating the major and minor axes directions. This is one of the ways that the additional degree of freedom manifests: if we fix the major/minor axes to be parallel to the y- and z-axes, then this requirement will force $\gamma = 0$ in the general equation for the ellipse, which will allow us to completely specify the ellipse on each plane parallel to the yz-plane, and which will allow us to completely specify the cone.


Consider the expression

$$ 2 z^2 + y^2 + 2 \sin(3 x)zy = x^2 $$

Here is a Mathematica Plot of that expression

enter image description here

Now I add the xz and xy planes, you clearly see the straight lines enter image description here

And now I show an arbitrary cross section (adding the plane $\{x = 0.7\}$) to see the ellipse

enter image description here

  • $\begingroup$ Very detailed answer. Thank you very much. In hindsight, I should have pointed out that the cone I was looking for didn't have an xy term in any of its traces, but your answer was very, very helpful. I thoroughly enjoyed reading it. On another note, for whatever reason, I can't seem to be able to see the pictures you uploaded. Either way, great answer. :) $\endgroup$
    – Hautdesert
    Aug 2, 2011 at 4:07
  • $\begingroup$ um, if you cannot see the pictures, please file a bug at the Meta site. (Go to meta.math.stackexchange.com ; ask a question, and tag it as (bug). Be sure to provide your browser and operating system information.) $\endgroup$ Aug 2, 2011 at 8:07
  • $\begingroup$ I can see them now. Nice. $\endgroup$
    – Hautdesert
    Aug 2, 2011 at 14:55

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