I am looking for the parameters/equation of an ellipse that would pass through all the vertices of an elongated parallelogram.
The ellispe has 5 free parameters: center $(x_0,y_0)$, major and minor axes $a$ and $b$ as well as an orientation $\theta$. The elongated parallelogram is fully defined by 3 vectors (6 parameters), therefore over-determining the system. However, there should be an analytical (and unique) solution to this problem !
Here is a visual help to understand my question (here, for simplicity, the three vectors are set at a 120° angle from each other):
Obviously, in some out-of-scope cases one might obtain a hyperbola instead of an ellipse. I would be very interested in a solution, even if one needs stronger assumptions (like the 120° angle between the vectors).
P.S.: The problem would probably be similar to finding the circumellipse of a triangle (orange), whose center (origin of the vectors) unfortunately doesn't match the basic Kimberling centers (incenter/centroid/etc.). The closest match I found is the Excentral-Hexyl Ellipse, but the center is not the right one.
Edit: The name elongated parallelogram actually refers to a centroid hexagon.
Opposite vertices of the hexagon have the same midpoint $O$, which is also the centre of the ellipse. Translate all vertices so that $O=(0,0)$: the equation of the ellipse simplifies then to $$ Ax^2+2Bxy+Cy^2=1. $$
Let $P$, $Q$, $R$ be any three non-opposite vertices of the hexagon (you can take for instance three consecutive vertices, or three alternate vertices as in your figure). Plugging the coordinates of those points into the above equation we can find $A,B,C$ (please check the result): $$ A={x_P y_P (y_Q^2 - y_R^2) + x_Q y_Q (y_R^2-y_P^2 )+x_R y_R(y_P^2 - y_Q^2) \over( x_P y_Q-x_Q y_P )(x_Q y_R - x_R y_Q )(x_P y_R-x_R y_P )}\\ B={1\over2}{x_R^2 (y_Q^2-y_P^2) + x_Q^2 (y_P^2 - y_R^2) + x_P^2 (y_R^2-y_Q^2 ) \over ( x_P y_Q-x_Q y_P )(x_Q y_R - x_R y_Q )(x_P y_R-x_R y_P )} \\ C={x_P y_P (x_Q^2 - x_R^2) + x_Q y_Q (x_R^2-x_P^2 )+x_R y_R(x_P^2 - x_Q^2) \over( x_P y_Q-x_Q y_P )(x_Q y_R - x_R y_Q )(x_P y_R-x_R y_P )} $$ Finally, the axes can be found from eigenvalues and eigenvectors of matrix $m=\pmatrix{A & B \\ B & C}$: eigenvalues are $1/a^2$ and $1/b^2$, while eigenvectors give the directions of the axes.