Intersection between elliptical parallel surface and line Geometric situation
Given is a surface in $\mathbb{R}^3$ that is equidistant to an ellipse. There is no special name for this type of parallel surface but if the ellipse becomes a circle we get a torus. The surface can be parametrically defined by
$$
\begin{align}
    \vec{x} 
=  \begin{bmatrix}
a\cos u-b r \cos u\cos v \ \left(b^2\cos^2 u+a^2\sin^2 u\right)^{-\frac{1}{2}} \\
b\sin u- a r \sin u\cos v\ \left(b^2\cos^2 u+a^2\sin^2 u\right)^{-\frac{1}{2}}\\
r\sin v
\end{bmatrix}
\end{align}
\tag{1}$$
where $a$ and $b$ are the known ellipse's semiaxes, and $r$ is the known radius of the circular cross-section, with $r<(a,b)$. The free parameters $u,v$ can be limited to the range $[0,2\pi]$. There is also a line with known parameters $(m_i,n_i)_{i=1,2,3}$ and free parameter $t$ in the range $[-\infty,+\infty]$ with equation
$$
\begin{align}
    \vec{y} 
=  \begin{bmatrix}
m_1\\
m_2\\
m_3
\end{bmatrix}
+  t\begin{bmatrix}
n_1\\
n_2\\
n_3
\end{bmatrix}
\end{align}
\tag{2}$$
Here's a plot of a possible situation (case 4 in the table below) where the line intersects twice the surface:

Mathematica Code of the plot:
a = 3; b = 1; r = 0.3;
s1 = ParametricPlot3D[{a Cos[u] - b r Cos[u] Cos[v]/Sqrt[b^2 Cos[u]^2+ a^2 Sin[u]^2], 
Sin[u] - a r Sin[u] Cos[v]/Sqrt[b^2 Cos[u]^2 + a^2 Sin[u]^2],
r Sin[v]}, {u, 0, 2 \[Pi]}, {v, 0, 2 \[Pi]}, Mesh -> 40];
s2 = ParametricPlot3D[{-3, 0, 0} + {-1, 0.2, -1} t, {t, -20, 10},
PlotStyle -> {Black,Thick}];
Show[{s1, s2}, PlotRange -> {{-4, 3}, {-3, 3}, {-4, 2}}, Axes -> None,Boxed -> False]

Question
There are 6 different intersections scenarios.
$\small{\begin{array}{c|c|c|c}
\text{Case} & \text{tangents} & \text{secants} & \text{sum} \\ \hline
1    & 0        & 0       & 0   \\ 
2    & 1        & 0       & 1   \\
3    & 2        & 0       & 2   \\
4    & 0        & 2       & 2   \\
5    & 1        & 2       & 3   \\
6    & 0        & 4       & 4  
\end{array}}$
Task: Check only if the number of intersections is 0 or above 0, i.e. do we have case 1 or not?
I tried to find if there are one or several combinations of $u,v,t$ so that $\vec{x}=\vec{y}$ but got only very complicated expressions. Is there a closed expression or do I have to use iterative methods?
Related question:
Shape defined by equidistant points to an ellipse in $\Bbb{R}^3$
 A: Consider the original ellipse $\mathbf{p}(u)$ that the "elliptical torus" is parallel to.
$\mathbf{p}(u) = (a \cos u , b \sin u , 0) $
and consider also the line $\mathbf{q}(t)$,
$\mathbf{q}(t) = \mathbf{m} + t \mathbf{n}$
Find the distance between $\mathbf{p}$ and $\mathbf{q}$, then minimize it.
If the minimum is less than $r$, then you have at least $2$ intersections. If the minimum is greater than $r$, then there are no intersections.
To minimize the distance, consider the square of the distance,
$f (u, t) = (\mathbf{p}(u) - \mathbf{q}(t))^T (\mathbf{p}(u) - \mathbf{q}(t))$
At the minimum of $f$, we have $\dfrac{\partial f}{\partial u} = \dfrac{\partial f }{\partial t } = 0 $
Thus, we have the following two equations:
$(\mathbf{p}(u) - \mathbf{q}(t))^T (- a \sin u ,  b \cos u, 0 ) = 0$
and
$(\mathbf{p}(u) - \mathbf{q}(t))^T (\mathbf{n} ) = 0$
These two equations can be solved for $u$ and $t$ using a numerical iteration such as the multivariate Newton-Raphson method, and it should converge to the solution very quickly (usually within 5 iterations).  Once you have $u$ and $t$ you can calculate the distance, and compare it with the radius $r$.
