We also search for the one with the larger area.
First of all, the obvious symmetry of the figure implies the ellipse center is at the rectangle center.
To formally prove that, just consider the two parallel tangents being two opposite sides.
The line connecting the two contact points is the polar of point at infinity on these tangents.
The ellipse center is the midpoint of this chord, on the median connecting the two other sides.
And similarily for the two other sides.
Also recall the known properties of ellipses : focii, director circle, circular directrix, properties of tangents...
Remains now to construct the focii and axes.
The axes may be constructed by considering the following pairs of conjugate diameters :
The focii are obtained as harmonic conjugates of intersections X and X' of the tangent and the normal in P
with the focal axis,
that is from OF² = OF'² = OX.OX' (again a classical construction)
The director circle is obtained from the classical property of the projection H of focus on the tangent being on
that director circle.
The ellipse vertices are immediately deduced, then the drawing friom a stretch of the director circle in ratio OV/OU.
In the applet, A and C define the rectangle ABCD. C is contrained by AB > BC > 0
P defines the contact point with AB
During the construction of axes, w choses the "any circle through O".
The previous construction of OF² = OF'² = OX.OX' fails if the focal axis is parallel to AB, that is when P is in M.
The applet prevents this case by restricting P in the segment MB - ε.
Note : The previous construction doesn't uses the angles of the rectangle ABCD being 90°, It is then valid without any change for a parallelogram ABCD.