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.

If "blocks" means "is circumscribed to" then you can find the answer here: https://math.stackexchange.com/questions/2928820/how-to-calculate-the-bounding-square-of-an-ellipse/2928831#2928831 — Intelligenti pauca, Jun 29 '20 at 10:05
Does this answer your question? how to calculate the bounding square of an ellipse? — Intelligenti pauca, Jun 29 '20 at 10:08
I think you are supposed to complete the answer with a bit of work. — Intelligenti pauca, Jun 29 '20 at 10:14
Vertices $(\pm u,\pm v)$ don't form a square unless $u=v$. So the bounding square of your ellipse can't have sides parallel to the axes. — TonyK, Jun 29 '20 at 10:36

Intelligenti pauca · Accepted Answer · 2020-06-29T11:04:58.510

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

Calculating the vertex of a square that circumscribed ellipse

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