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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.

Rushabh Mehta
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anonymous
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  • 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... – Jean Marie Mar 14 '21 at 18:53
  • 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. – Jean Marie Mar 14 '21 at 21:32
  • See if this can be of help: https://math.stackexchange.com/questions/2651714/from-ellipse-equation-to-circular-cone-axis/2651961#2651961 – Intelligenti pauca Mar 14 '21 at 21:53
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    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) – Blue Mar 15 '21 at 02:20
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    (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. – Blue Mar 15 '21 at 02:21
  • I know how to get ellipses with all eccentricity. If I keep fixed the point of intersection of the plane and the axis of the conic and move the plane in such a manner that the inclination angle runs from $V$ to $\frac{\pi}{2}$ , we will get ellipses with all possible eccentricity. But my question is how can I get an ellipse with the major axis of length $d$. Is there any method we can do so ?@Blue – anonymous Mar 15 '21 at 02:56
  • I think the eccentricity will be $\frac{\cos U}{\cos V}$ not $\frac{\sin U}{\sin V}$ as V is fixed and we are varying the angle U. When U>V , we should get ellipse ; that is $e<1$.@Blue – anonymous Mar 15 '21 at 04:02

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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 $R$.

enter image description here

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.

Blue
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