What are the axes of an ellipse within a rhombus?

Question

For a rhombus of x width and y height, what are the major and minor axes of the unique inscribed ellipse tangent to the midpoints of the rhombus's sides? Thank you.

Answer 1

Let $4a$ be the longest diagonal, and $4b$ be the shortest one (see figure with $a=2$, $b=\frac{3}{2}.)$

It is not difficult to show (see explanations below) that the equation of the inscribed ellipse in the rhombus that is tangent to it in the midpoints of its sides is:

$$\tag{1}\frac{b}{a}x^2+\frac{a}{b}y^2=2ab$$

Putting this in the equivalent standard form for the equation of an ellipse having coord. axes as its symmetry axes:

$$\frac{x^2}{(a\sqrt{2})^2}+\frac{y^2}{(b\sqrt{2})^2}=1$$

meaning that the looked-for half-axis are

$$A=a\sqrt{2} \ \ \text{and} \ \ B=b\sqrt{2}.$$

See figure below with $2a=4$, $2b=3$, $A=2\sqrt{2}\approx 2.83..$, $B=\frac{3}{2}\sqrt{2}\approx 2.12.$

Explanations for (1): First note the - expected - symmetry in $a$ and $b$.

Equation (1) is verified by midpoint $E(x=a, y=b)$ and more generaly by the four points $(x=\pm a, y=\pm b)$ (see figure).

Besides, the tangent equation to the ellipse in $(x_0,y_0)$ is

$$\tag{2}\frac{b}{a}xx_0+\frac{a}{b}yy_0=2ab.$$

Taking $x_0=a$, $y_0=b$ in (2) gives the equation $\frac{x}{a}+ \frac{y}{b}=1$ of straight line CD, as awaited.

Answer 2

As mentioned by others, a given rhombus admits infinitely-many inscribed ellipses. Perhaps unexpectedly, there's an interesting way to construct the family. First, some algebra:

Suppose an origin-centered rhombus with vertices $P=(p,0)$ and $Q=(0,q)$ has an origin-centered (and axis-aligned) inscribed ellipse through $A = (a,0)$ and $B=(0,b)$. As line $\overleftrightarrow{PQ}$ only touches the ellipse once, there can be at exactly one common solution to their equations $$\frac{x}{p} + \frac{y}{q} = 1 \qquad\text{and}\qquad \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ Eliminating $y$ from the above, and solving for $x$, gives $$ x = \frac{ap}{a^2q^2+b^2p^2}\left(a q^2 \pm b \sqrt{ a^2 q^2 + b^2 p^2 - p^2 q^2} \right)$$ but, since there should be exactly one solution, we conclude that the discriminant must vanish:

$$a^2 q^2 + b^2 p^2 - p^2 q^2 = 0 \qquad\to\qquad \frac{a^2}{p^2} + \frac{b^2}{q^2} = 1 \tag{$\star$}$$

Observe that the second form of the equation in $(\star)$ says precisely that the point $K=(a,b)$ lies on the (axis-aligned) ellipse through the vertices of the rhombus:

Indeed, every point on the circumscribed ellipse gives rise to an inscribed ellipse:

Neat!

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Edit. If, as OP later stated, the question is specifically about the ellipse tangent to the midpoints of the sides of the rhombus, then the solution is easy.

We leverage the fact that tangency, midpoint-ness, and ellipse-ness are affinely-independent notions. That is to say, they're preserved under various affine transformations, in particular: scaling (aka stretching). The upshot here: If we can solve this problem for a square, then we've solved the problem for all rhombi. Of course, solving the problem for a square is easy:

For a square with diagonal-length $2s$, the radius of the circle inscribed in that square (and necessarily tangent at the midpoints of its sides) is clearly $\frac{\sqrt{2}}{2} s$. If we stretch the square into a rhombus with diagonal-lengths $2p$ and $2q$ ...

... then we correspondingly scale "horizontal" distances by $p/s$ and "vertical" distances by $q/s$. Consequently, the radii of the inscribed circle-turned-ellipse (which remains tangent at the midpoints of the sides of the square-turned-rhombus) are

$$a = r \cdot \frac{p}{s} = \frac{\sqrt{2}}{2}s\cdot\frac{p}{s} = \frac{\sqrt{2}}{2} p \qquad\text{and}\qquad b = \frac{\sqrt{2}}{2}q$$