Calculating the vertex of a square that circumscribed ellipse

Question

How can I find the vertex of a square that circumscribes the ellipse defined as follows?

$$\frac{x^2}{9}+y^2 =1$$

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.

Answer 1

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 $(\pm\sqrt{10},0)$.