Hint: Solve the non-linear system:
This will give you two opposing vertices, from which you may determine the remaining two by symmetry.
From there you should have little difficulty in determining the area and perimeter of the square.
Hint: Solve the non-linear system:
This will give you two opposing vertices, from which you may determine the remaining two by symmetry.
From there you should have little difficulty in determining the area and perimeter of the square.
As ellipse is symmetric about the origin and square being a cyclic quadrilateral that is symmetric about the origin with all side equal the only possibility is as show in the figure.
Let the side of the square be so the vertices are substituting in the equation of the ellipse we have:
the vertices are . The rest can be determined.