how for any $(a , e)$ we will get an ellipse on the double cone by the intersection of a Plane and the cone 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.
 A: 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 $R$.

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.
