An ellipse is specified in vector form as follows
$P_1(t) = C_1 + V_1 \cos(t) + V_2 \sin(t) $
where $C_1$ is the center of ellipse, and $V_1, V_2$ are mutually orthogonal, and extend along the semi-major and semi-minor axes of the ellipse. And a hyperbola is described in vector form by
$P_2(s) = C_2 + W_1 \sec(s) + W_2 \tan(s) $
where $C_2 $ is the vertex of the parabola, and $W_1, W_2$ are mutually orthogonal, and $W_1$ extends along the major axis of the hyperbola, while $W_2$ extends along the minor axis.
Question: Find the minimum distance between the ellipse and the hyperbola.
My attempt:
My attempt is described in my solution that follows.
Since at the minimum distance, the vector extending from the respective points on the ellipse and the hyperbola $(P_1 P_2)$ is perpendicular to the ellipse and to the hyperbola, then we want
$ (P_1 - P_2) \cdot P'_1 = 0 $
$ (P_1 - P_2) \cdot P'_2 = 0 $
where $P'_1$ is the tangent vector to the ellipse, and is given by
$ P'_1(t) = -V_1 \sin(t) + V_2 \cos(t) $
Similarly,
$ P'_2(s) = W_1 \sec(s) \tan(s) + 2 W_2 \sec^2(s) $
Using the Newton-Raphson multivariate method, in the two variables $t, s$, solutions to these equations can found, and the distance at each pair evaluated and compared, resulting the minimum possible distances.
Example:
Let's take
$C_1 = (0, 10) , V_1 = (-2, 1) , V_2 = (3, 6) $ for the ellipse, and
$C_2 = (-5, -2), W_1 = (\sqrt{5},0) , W_2 = (0, \sqrt{\dfrac{5}{3}}) $ for the hyperbola
We get a minimum distance (indicated by the green line segment) of $d_{Min} \approx 3.996541 $
This is shown in the figure below.