Find minimum-area ellipse which encloses two ellipses I need an efficient algorithm to find the ellipse with the smallest possible area which encloses two given ellipses. The given ellipses are constrained to have coincident centers at the origin but can have any orientation. The problem is limited to two dimensions. Any ideas?
 A: Yeah, there is a shear transformation that takes one of the ellipses to a circle. The least area ellipse enclosing the resulting figure is now evident by symmetry. Then use the inverse of the shear.
Now that I think of it, you can just shrink along the major axis of one of the ellipses and expand on the minor axis to get the circle.
All manipulations involved are with 2 by 2 matrices. The hypothesis of coincident centers is crucial.
A: Below is an example to illustrate Will Jagy's solution.
(Caveat lector: I did not preserve scale from image to image.)
First, it is no loss of generality to rotate so that one ellipse $E_1$ has its major
axis along the $x$-axis:

          


Now transform by scaling the long axis down and the short axis up so that
$E_1$ becomes a circle.  The determinant of this transformation matrix is $1$, so areas are preserved. The second ellipse $E_2$ is transformed to another ellipse:

          


Compute the major axis of the transformed $E_2$:

          


Rotate $E_2$ down so its major axis is along the $x$-axes:

          


Now it is clear what the minimum area enclosing ellipse is, as its two axes are determined
by the ellipse and the circle:

          


Rotate back, and unscale:

          



Here is Will Jagy's drawing as he mentions in his comment below:

          


