I understand Dandelin spheres.

But I can not understand how for any $(a , e)$ we will get an ellipse on the double cone by the intersection of a Plane and the cone.

Here $a$ is the half of the major axis and $e$ is the eccentricity.

  • $\begingroup$ Do you mean a "cone of revolution" ? Because the word "cone" can also designate a "broader" family... In such a case, it would be sufficient to place your ellipse anywhere on the horizontal plane, take an arbitrary point $S$ with a $z \ne 0$ and connect $S$ to all points of the ellipse... $\endgroup$
    – Jean Marie
    Mar 14, 2021 at 18:53
  • $\begingroup$ You can reduce your investigation to a planar section of the (looked for) cone along the semimajor axis of the ellipse. As you know the Dandelin spheres, you could find this answer interesting. $\endgroup$
    – Jean Marie
    Mar 14, 2021 at 21:32
  • $\begingroup$ See if this can be of help: math.stackexchange.com/questions/2651714/… $\endgroup$ Mar 14, 2021 at 21:53
  • 1
    $\begingroup$ As mentioned in this answer (which you've read), eccentricity is the ratio $e=\sin U/\sin V$ (where $U$ is the angle of inclination of the cutting plane, and $V$ is the angle made by (the generators of) the surface of the cone). Varying $U$ from $0$ to $V$ gives every eccentricity from $0$ (circle) to $1$ (parabola), so every elliptic shape occurs somewhere. To account for every size, we vary how far the cutting plane's intersection with the axis is from the vertex; everything "scales" to get every $a$ from $0$ to $\infty$. (continued) $\endgroup$
    – Blue
    Mar 15, 2021 at 2:20
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    $\begingroup$ (continuing) To get hyperbolas, one uses a cutting plane that's steeper than the cone (so that the plane cuts both nappes of the cone); that is, $U>V$ and likewise $\sin U> \sin V$, making $e=\sin U/\sin V> 1$. Interestingly, since $\sin U\leq 1$, the largest eccentricity achievable with an angle-$V$ cone is $1/\sin V$. So, although every ellipse, and the parabola, appears as a section of a given cone, not every hyperbola does. To get all hyperbolas, you need to vary the cone's angle as well. $\endgroup$
    – Blue
    Mar 15, 2021 at 2:21

1 Answer 1


Expanding-upon a comment ...

As mentioned in this answer the eccentricity of a conic is given by $$e = \frac{\sin U}{\sin V} \tag1$$ where $U$ is the angle of inclination of the cutting plane, and $V$ is the angle of (the generators of) the cone.*

By varying $U$ from $O$ to $V$, the value of $e$ varies from $0$ (circle) to $1$ (parabola), so that every ellipse shape is achieved by simply tilting the plane.

To achieve every possible size, we vary the distance from the cutting plane to the cone's vertex. For specificity, consider this "side view" of the situation, with the cutting plane meeting the cone's axis at $P$, and the vertices of the resulting ellipse are $Q$ and $R$.

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The Law of Sines tells us that, in $\triangle PVQ$, $$\frac{q}{\sin\angle PVQ}=\frac{p}{\sin\angle PQV} \quad\to\quad \frac{q}{\sin(90^\circ-V)}=\frac{p}{\sin(V+U)} \quad\to\quad q = \frac{p\cos V}{\sin(V+U)} \tag2$$ and likewise, in $\triangle PVR$, $$\frac{r}{\sin\angle PVR}=\frac{p}{\sin\angle PRV} \quad\to\quad \frac{r}{\sin(90^\circ-V)}=\frac{p}{\sin(V-U)} \quad\to\quad r = \frac{p\cos V}{\sin(V-U)} \tag3$$

Consequently, $|QR|$, the major axis of the ellipse, is given by $$|QR| = q+r = \frac{p\cos V(\sin(V-U)+\sin(V+U))}{\sin(V-U)\sin(V+U)} = \frac{p\cos U\sin 2V}{\sin(V-U)\sin(V+U)} \tag4$$

(Sanity check: When $U=0$, this reduces to $2p\cot V$, the width of the cone at the level of $P$, as we would expect with a circle. When $U=V$, $\sin(V-U)=0$ so that the length becomes infinite, as we would expect with a parabola.)

So, for a given pair of angles $U$ and $V$, we can choose $p$ so that $(4)$ yields whatever major axis length we desire.

*These angles are measured against "the horizontal". If instead one chooses to measure against, say, the "vertical" axis of the cone, then the ratio would be $\cos U/\cos V$.

I prefer sines, as they make zero-values correspond to zero-angles, so that in particular a circle ($e=0$) corresponds to a horizontal cutting plane with $0^\circ$ inclination.


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