An ellipse tangent to a symmetric trapezoid

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First note that the figure has a reflective symmetry; it is preserved after reflecting about the line connecting the midpoints \(M_1\) and \(M_2\) of the top side and bottom side of the trapezoid. Therefore the minor axis of the ellipse is the segment \(M_1 M_2\).
We also know that the major axis is perpendicular to the minor axis and it passes through the midpoint. Therefore we can construct the line of the major axis as the perpendicular bisector of \(M_1 M_2\) and the problem reduces to finding the length of the major axis.
Recall the Concentric Circle Construction of the ellipse. In this problem we turn the construction around and use our knowledge of the minor axis and the given points on the ellipse to construct the outer circle in the concentric circle construction. This then tells us the length of the major axis.
Draw a circle centred at the intersection \(O\) of the major and minor axes, with radius equal to half the length of the minor axis. This is the inner circle from the concentric circle construction. From one of the points \(A\) on the ellipse, draw a line parallel to the major axis and mark the point \(X\) where this intersects the circle just drawn. Draw a ray from \(O\) to \(X\) and let \(Y\) be the point where this ray intersects a line through \(A\) parallel to the minor axis. Then \(Y\) lies on the outer circle in the concentric circle construction, and so the length \(|OY|\) is the radius of this circle, which is equal to half of the length of the major axis.
Now that we know the length of the major axis we can construct the focal points and the ellipse.