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Like this:

enter image description here

where inner the angles of this rhombus are 60° and 120°, and the sides have length 1.

We can see that the sides are tangent to the inscribed ellipse at their midpoints.

I believe that the two intersections of the perpendiculars of the sides shown here are the foci of the ellipse… but I can find no confirmation of that anywhere. If we can assume that is the case, then I worked out that the ellipse’s minor axis has length $\sqrt{\frac{1 + \sqrt 5}{6}}$ and the major axis is $\frac{1 + \sqrt 5 }{2 \sqrt 3} $.

EDIT: Nope! Bad assumption. See below.

So a related question would be, are these really the foci of this ellipse? If not, then how to construct them properly?

I’ve looked everywhere for these answers, to no avail. I have found many videos and websites explaining how to approximate this ellipse using a compass, a technique which makes use of the pair of points that I’m assuming are foci. It seems like the only people interesting in this are draftsmen and artists, not mathematicians. Quite frustrating and disappointing!

A bit more discussion here: https://www.quora.com/What-are-the-lengths-of-the-major-and-minor-axes-of-an-ellipse-inscribed-in-a-rhombus-of-an-isometric-grid-i-e-with-opposite-angles-of-60-and-120

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    $\begingroup$ Your rhombus is the result of stretching a square by a factor of $\sqrt{3}$ along one diagonal. Consequently, your midpoint-tangent ellipse is the result of stretching that square's inscribed circle along with it. Can you take things from here? $\endgroup$
    – Blue
    Nov 21 '20 at 10:09
  • $\begingroup$ Those axis lengths are wrong (see my answer), so it seems that your conjecture about the foci must be wrong too. $\endgroup$
    – TonyK
    Nov 21 '20 at 16:10
  • $\begingroup$ yep, totally wrong. it tuns out the foci simply lie at $\pm \frac{1}{2}$ $\endgroup$
    – Marc
    Nov 21 '20 at 16:13
  • $\begingroup$ This means that the major axis has length $\frac{\sqrt{6}}{2}$ and the minor axis $\frac{\sqrt{2}}{2}$ $\endgroup$
    – Marc
    Nov 21 '20 at 16:19
  • $\begingroup$ Just to be clear: there are an infinite number of ellipses that can be inscribed in your rhombus. You have to specify that the ellipse meets the sides of the rhombus at their midpoints; it is not something that you can deduce. $\endgroup$
    – TonyK
    Nov 22 '20 at 21:29
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@Blue's comment shows you the way to go. Here is what your diagram looks like, squeezed horizontally by a factor of $\frac{1}{\sqrt 3}$:

enter image description here

Your rhombus with unit side lengths has become a square of side $\frac{1}{\sqrt2}$, so your ellipse has become a circle with diameter $\frac{1}{\sqrt2}$. Now stretch it back out; the minor axis remains unchanged, and the major axis increases by a factor of $\sqrt3$.

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