Take a circle. The largest-area inscribed triangle will be an equilateral triangle. The proportion of the circle's area covered by the inscribed triangle will not change when that circle is flattened into an ellipse, as if seen from an angle other than head on (in orthographic perspective, rather than "perspective perspective" -- an affine transformation).

What is the area of an equilateral triangle inscribed in a circle, relative to the circle's area?

A circle with radius 1 will have an area equal to pi.

One can consider the inscribed equilateral triangle to be three isosceles triangles with apex angles equal to 120° meeting at the center. Each of these can be halved, resulting in six right triangles with hypotenuse 1 and a 60° angle. Each of the triangles has sides of 1/2, 1, sqrt(3)/2, for an area of sqrt(3)/8. There are six of them, so the total for the entire equilateral triangle is 3*sqrt(3)/4.

The fraction of the circle's area that is also the triangle's area is 3*sqrt(3)/(4*pi). That will also be the fraction of the ellipse's area covered by the maximal inscribed triangle.

The major and minor semiaxes of the ellipse are 5 and 4 respectively, so the ellipse has area 20*pi. Multiplying by the fraction included within the maximal inscribed triangle, we get 15*sqrt(3).