Calculating The Vertex Of a Square That Circumscribed Ellipse Hello everyone how can I find the vertex of a square that circumscribed the ellipse $\frac{x^2}{9}+y^2 =1$?
I tried to mark the vertex at $(u,v),(-u,v),(u,-v),(-u,-v)$ and use the equation to calculate the tangent lines to the ellipse by the vertex points, but I don't know how to continue.

It is possible to give a very quick answer if we use this nice property of the ellipse:

the locus of the intersections of perpendicular tangents to an
ellipse is a circle called director circle,
and the square of its radius is the sum of the squares of
the ellipse semi-axes.

Tangents drawn from any point on the director circle, and from its reflection about the center, form then the sides of a circumscribed rectangle. If we choose those points as the intersections between director circle and axes of the ellipse, the rectangle is by symmetry a square.
In your particular case the vertices of the circumscribed square lie then at points $(0,\pm\sqrt{10})$ and $(\pm\sqrt{10},0)$.
