Ellipses and circles are "really" the same: a circle is an ellipse in which the focal distance is $0$. Or, put another way, from among all the ellipses
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$
if you are going to choose a "canonical" one to define some functions, the natural choice is to use $a=b=1$, which just leads you back to the unit circle. That is, trying to do "elliptic trigonometric functions" pretty soon either drops you into either an arbitrary choice of parameters, or the regular circular trigonometric functions.
Now, one way in which we can "unify" the circular and the hyperbolic functions is through the complex exponential: the circular trigonometric functions correspond to certain exponentials with purely imaginary arguments, while the hyperbolic ones correspond to purely real arguments: for $t$ a real number,
$$\begin{align*}
\cos t&= \frac{e^{it}+e^{-it}}{2} &\qquad \sin t &= \frac{e^{it}-e^{-it}}{2}\\
\cosh t &= \frac{{e^t}+e^{-t}}{2} & \sinh t &= \frac{e^t-e^{-t}}{2}
\end{align*}$$
Where do these come from? One place to find them is the differential equation
$$y'' + \lambda y = 0.$$
If $\lambda\gt 0$, the solution set is spanned by $\cos(\sqrt{\lambda}\;t)$ and $\sin(\sqrt{\lambda}\;t)$. You get the standard functions by normalizing with $\lambda=1$. If $\lambda\lt 0$, then the solution set is spanned by $\cosh(\sqrt{|\lambda|t})$ and $\sinh(\sqrt{|\lambda|}t)$, and you get the standard functions by normalizing with $\lambda=-1$.
This would suggest looking for any putative "parabolic trigonometric functions" in the only remaining case: $\lambda=0$ (this corresponds to the fact that if you view conic sections as being given by slicing a cone with a plane, you obtain the parabola in the boundary between ellipses and hyperbolas). But the differential equation $y''=0$ has solution space spanned by $1$ and $t$, so that the natural "parabolic functions" are just $1$ and $t$.
Converting comments (about the original version of the question) to answer, as requested.
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.
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
$$\frac{r'}{r} = \frac{\tan\frac12(\alpha+\beta)}{\tan\frac12(\alpha-\beta)} \tag{3}$$
Best Answer
Leaving aside the presence of imaginary ellipses, what you ask is the probability that, given three numbers $a$, $b$, $c$ chosen at random in $[-1,1]$, then $b^2>4ac$ (hyperbolic case).
But $b^2=4ac$ is the equation of a cone in $abc$ space, hence that probability is the same as the fraction of the volume of cube $[-1,1]^3$ which is outside that cone. To compute this volume, let's focus on the first octant and unit cube $[0,1]^3$. The intersection between the cone and a plane $b=\text{constant}$ is a hyperbola in $ac$-plane with equation $ac=b^2/4$. The intersection of the same plane with the unit cube is a square and the area in that square outside the hyperbola is: $$ A(b)={b^2\over4}+\int_{b^2/4}^1{b^2/4\over a}da= {b^2\over4}\left(1-\ln{b^2\over4}\right). $$ The required volume can then be computed by integrating that over $b$: $$ V=\int_0^1 A(b)\,db={5\over36}+{1\over6}\ln2. $$ The cone intersects the original $[-1,1]^3$ cube in other three octants, with the same external volume, while the unit cubes in the remaining four octants are completely outside the cone. Hence the requested probability is: $$ p={1\over8}(4V+4)={41\over72}+{1\over12}\ln2\approx 0.627207. $$ This result is confirmed by a quick simulation with Mathematica: