This time I have all three rectangles parallel to each other but not joined in one big brick. This time all three rectangle centers will remain on the x-axis.
Like last time, my ellipse is x^2/9+y^2=1. The 'vertical' sides of the rectangles tangent to the ellipse are ((cos t)/3)*x + (sin t)*y = +/-1. The 'vertical' sides of the center rectangle are ((cos t)/3)*x + (sin t)*y = +/-1/3. (I use the word vertical loosely here to represent the sides/lines I expect to be closer to vertical slope than horizontal slope in the solution. At t=0 these are true vertical lines.)
The horizontal lines of the center rectangle are (sin t)*x - ((cos t)/3)*y = 1.
At t=0 the vertecies are far outside the ellipse, so the answer is a matter of finding the angle when a vertex lies on the ellipse. Again, I opted for the spreadsheet to get a value of t=0.24261262721
The vertecies of the center rectangle are at (1.65021370961, -0.83511773705); (-0.32203403613, 1.82124160992);
(-1.65021370961, 0.83511773705); (0.32203403613, -1.82124160992)
The sides of each rectangle are 3.30847489802 by 1.65423744901, for an area of 5.47300307540 for each rectangle. This time the ratio of ellipse to total rectangle area is 0.57401624123, an improvement over the simple big rectangle.