## Minimal ellipse circumscribed on a parallelogram

Consider a parallelogram p = ABCD. Of all the ellipses circumscribed about the parallelogram, there is one (m), which has minimum area. The directions of the diagonals of the parallelogram are conjugate directions of that ellipse.

The proof reduces to the analogous proposition for ellipses circumscribed about a square. It is well known, that the circumcircle is the minimum (in area) circumscribed ellipse on the square. Look at the file MinimumEllipse.html for a short proof. By the general properties of projectivities, there is exactly one projectivity F, mapping the vertices of a square A', B', C', D' to the vertices of the parallelogram A, B, C, D, correspondingly. Since, in that case the projectivity maps two points at infinity, corresponding to the pairs of parallel sides of the square, to two points at infinity, corresponding to the two pairs of parallel sides of the parallelogram, the map leaves invariant the line at infinity hence is an affinity.
The image of the square's circumcircle under this affinity is the minimal circumscribed ellipse (in area) m of the parallelogram. In fact, the affinity F multiplies the area of shapes by a number (the determinant of the matrix, representing the afinity). i.e. if c is a circumellipse of the square and F(c), its image, a circumellipse of the parallelogram, then the ratio Area(F(c))/Area(c) = k, is a constant. This implies trivially the statement.
Let now (c) be the circumcircle of the square. Orthogonal directions, from the center of the square, map under F, to conjugate directions of the ellipse F(c). This follows from the fact that F preserves parallel lines, hence a direction (IK) maps to (JM) and its parallel tangent (LQ), maps to a tangent (NP), parallel to (JM). Thus (IL), orthogonal to (IK) maps to (JN), which is conjugate to (JM). The statement on the diagonals of the parallelogram, follows by taking K to coincide with a vertex of the square.
Look at the file ParaDivision.html for a nice application to the problem of dividing the parallelogram in four equal parts through two intersecting lines.
Consider all ellipses inscribed in the parallelogram, there is one which has maximal area. The directions of the diagonals of the parallelogram are conjugate directions of that ellipse. Give a proof of that analogous to the previous one. Look at the file MaximalEllipse.html for the case of ellipses inscribed in a square. Show in addition that the two ellipses circumscribed/inscribed in the same parallelogram p are homothetic with ratio sqrt(2).

MaximalEllipse.html
MedialParabola.html
MinimumEllipse.html
Miquel_Point.html
Parabola.html
ParabolaProperty.html