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
$P_2(s) = C_2 + W_1 s + W_2 s^2 $
where $C_2 $ is the vertex of the parabola, and $W_1, W_2$ are mutually orthogonal, and $W_2$ extends along the axis of the parabola.
Question: Find the minimum distance between the ellipse and the parabola.
My attempt: Since at the minimum distance, the vector extending from the respective points on the ellipse and the parabola $(P_1 P_2)$ is perpendicular to the ellipse and to the parabola, 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 + 2 W_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, 15) , V_1 = (3, 1) , V_2 = (-2, 6) $ for the ellipse, and
$C_2 = (8, 0), W_1 = (1,-1) , W_2 = (0.25, 0.25) $ for the parabola
We get a minimum distance (indicated by the green line segment) of $d_{Min} \approx 5.253056 $
This is shown in the figure below.
