Dandelin sphere construction with hyperbola and ellipse intersection on same cone

Question

When eccentricity of ellipse $\epsilon<1 $ sketch is often made in a standard disposition during proof by Dandelin with spheres inside a single nappe of cone and cutting plane touching spheres in their central section at foci of ellipses so created.

But how do we construct this ambiance and calculate a hyperbola conic intersection $\epsilon >1 $ in association with the Dandelin spheres? Is my following sketch correct?

EDIT1:

The error has since been corrected, duality noted. But for time being cone is not changed but variable sectioning is done.

$$\text{ Inverse tangents Cone Meridians for Ellipses $\epsilon <1 $ } $$ $ (\gamma,\alpha_E)$relations are $$\sin \gamma = (R - r)/d \,;\, \,\sin \alpha_E = (R+ r)/d;\, \gamma< \alpha \tag{1a}$$ $$ \epsilon_{ellipse} = \frac{\cos \alpha_E}{\cos\gamma}=\sqrt{\frac{(d + R + r) (d - R - r)}{((d - R + r) (d + R - r)}},\;\, d>(R+r) \tag{2a} $$

$$\text{ Direct tangents Cone Meridians for Hyperbolas $\epsilon >1$ shown } $$ $ (\gamma,\alpha)$relations are $$\sin \alpha_H = (R - r)/d \,;\, \,\sin \gamma = (R+ r)/d;\, \gamma> \alpha \tag{1b}$$ $$ \epsilon_{hyperbola} = \frac{\cos \alpha_H}{\cos\gamma}=\sqrt{\frac{(d - R + r) (d + R - r)}{((d + R + r) (d - R - r)}},\;\, d>(R+r) \tag{2b} $$

It is seen that there is a simple swap of cone & cutting angles from ellipse to hyperbola intersections and vice versa. In fact interchanging their colors of what they represent (Cone&Cutting Plane) the same sketch would serve either case in a duality situation.

For calculation if we take data for ellipse

$$(R,r,d)=(5,3,10 ) \rightarrow \epsilon\approx 0.6124 $$

and for the " radius $r$ mirrored " case of same $(R,r,d)$ producing hyperbola intersections we have $(R,r,d)=(5,3,10 ) \rightarrow \epsilon\approx 1/ 0.6124 \approx 1.633 $

EDITS (2,3):

The basic question here is whether there are two cones or one cone for a combined representation of Dandelon spheres $(R,r)$ in elliptic/hyperbolic intersections.

In the sketch here I have fixed the cone ( its angle, position) anf with differently inclined cutting planes. The second ellipse intersection at left is for time being ignored. I wish to bring out the two cases where product of eccentricities is unity.

Unless both are seen together in a common configuration a comprehensive picture somehow suffers imho, so this question.

EDIT 4:

It takes two equal diameter Dandelin spheres contained separately in each nappe of a cone (say semi-vertical angle $45^{\circ}$) touching each nappe separately in order necessarily to result in an equiangular hyperbola intersection by a plane parallel to cone axis.

Answer 1

Converting comments (about the original version of the question) to answer, as requested.

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The cutting plane should be tangent to the Dandelin spheres. For the hyperbola case, looking from the side (as in your image), the "cutting line" should be externally tangent to the two "Dandelin circles".

It may be easier, in all cases, to start with the cone and cutting plane. Again, looking from the side, the "cone lines" and "cutting line" will determine, in the cone's interior, certain regions bounded by three sides (segments and/or rays). For an ellipse, there will be a finite triangle and an unbounded three-sided region; for a hyperbola, there will be two unbounded three-sided regions; for a parabola, there will be one unbounded three-sided region. Each "Dandelin circle" is tangent to the three sides of such a region.

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Here's "unified" picture of a Dandelin configuration for both ellipses and hyperbolas (and parabolas as a limiting case). It consists simply of two circles, $\bigcirc K$ and $\bigcirc K'$, and their common tangents: the internal tangents meet at $A$, making an angle $\alpha$ with the line of centers; and the external tangents meet at $B$, making an angle $\beta$ with the line of centers. Necessarily, $\alpha > \beta$. (Exploration of the case where $\alpha=\beta$, which defines a parabola, is left to the reader.)

As indicated, $V=W$, $V'$, and $W'$ are points where an internal tangent meets an external. Internal tangent $\overleftrightarrow{VV'}$ touches the circles at $F$ and $F'$; external tangent $\overleftrightarrow{WW'}$ touches the circles at $G$ and $G'$. Note: because $|VF'|=|WG'|$ and $|VF|=|WG|=|V'F'|=|W'G'|$ (verification left to the reader), we have $|VV'|=|GG'|$ and $|WW'|=|FF'|$.

This allows us to restate the calculations made by OP:

• For an ellipse, we interpret $\overleftrightarrow{VV'}$ as the "cutting line"; that is, (the side-view) of the cutting plane. And $\overleftrightarrow{WW'}$ is a "cone line", a generator of the cone. Here, $V$ and $V'$ are the vertices, while $F$ and $F'$ are the foci, and we have $$e = \frac{|FF'|}{|VV'|}=\frac{\color{red}{|FF'|}}{\color{blue}{|GG'|}}=\frac{|KK'|\cos\alpha}{|KK'|\cos\beta}= \frac{\cos\alpha}{\cos\beta}< 1 \tag{1}$$
• For an hyperbola, we interpret $\overleftrightarrow{WW'}$ as the "cutting line" and $\overleftrightarrow{VV'}$ as a "cone line". Here, $W$ and $W'$ are vertices, while $G$ and $G'$ are foci, and we have $$e = \frac{|GG'|}{|WW'|}=\frac{\color{blue}{|GG'|}}{\color{red}{|FF'|}}=\frac{\cos\beta}{\cos\alpha}> 1 \tag{2}$$

It's worth reiterating that $(1)$ and $(2)$ say exactly that eccentricity of a conic is simply a ratio of internal and external tangent segments for these "Dandelin circles". I don't believe this is a new or particularly-profound observation, but I personally have never thought of things in quite this way. Nifty!

Also, as mentioned by OP, we have these relations involving the Dandelin radii, $r := |KF|=|KG|$ and $r':=|K'F'|=|K'G'|$: $$r'+r = |KK'|\sin\alpha \qquad r'-r=|KK'|\sin\beta$$

These imply $$\begin{align} r &= \frac12|KK'|(\sin\alpha-\sin\beta) = |KK'| \cos\frac12(\alpha+\beta)\sin\frac12(\alpha-\beta) \\[4pt] r' &= \frac12|KK'|(\sin\alpha+\sin\beta) = |KK'| \sin\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta) \\ \end{align}$$ so that $$\frac{r'}{r} = \frac{\tan\frac12(\alpha+\beta)}{\tan\frac12(\alpha-\beta)} \tag{3}$$