$5$ points on ellipse

Question

I want to take five points and locate them on ellipse so that the distance between the point $1$ and point $2$, the point $2$ and point $3$, the point $3$ and point $4$, the point $4$ and point $5$, the point $5$ and point $1$ were equal to each other. I think it will look like a polygon with equal side lengths, but with different angles. Can you help me to figure out how to find the distance among those points and the coordinates of those points if the minor and major axes are $b$ and $a$ respectively?

Answer 1

In a fixed ellipse there are infinite inscribed equilateral pentagons, with different side lengths. You may start from a point $P$ on the ellipse and define, for any sufficiently small $\ell>0$, $f_{\ell}(P)$ as the point $Q$ of the ellipse such that $PQ=\ell$ and $P$ occurs "shortly before" $Q$ when travelling the ellipse counter-clockise. Let $R=(f_\ell\circ f_\ell\circ f_\ell\circ f_\ell\circ f_\ell)(P)$. For small values of $\ell$, $R$ occurs "shortly after" $P$, but for large values of $\ell$ the point $R$ occurs "shortly before" $P$. By continuity, for some value of $\ell$ (which depends on the location of $P$) we have $R\equiv P$, leading to an inscribed equilateral pentagon. But such function ($P\mapsto \ell_{\text{crit}}$) is not constant on the ellipse: in general we have infinite inscribed equilateral pentagons, with slightly (very slightly, if the eccentricity is small) different side lengths.

Hence a range of possible side lengths, not a single side length:

Answer 2

Equation of the ellipse : $$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1$$ with given values $a,b$.

They are many solutions, depending where we put the first point on the ellipse.

If you want only one solution, one of the simplest is obtained in putting the first point A on $(x=a,y=0)$.

See Figure below. With AB=BC=CD=DE=EA=$L$, at beginning $L$ is unknown. To find it, the equations are :

Point A$(x_1,y_1)$ : $$y_1=\sqrt{1-\left(\frac{x}{a}\right)^2} \quad;\quad y_1^2+(a-x_1)^2=L^2$$ $$x_1=a\frac{a^2-\sqrt{b^4+(a^2-b^2)L^2}}{a^2-b^2} \tag 1$$ $$y_1=b\sqrt{1-\left(\frac{a^2-\sqrt{b^4+(a^2-b^2)L^2}}{a^2-b^2}\right)^2} \tag 2$$ Point B$(x_2,y_2)$ : $$y_2=\frac{L}{2} \tag 3$$ $$x_2=a\sqrt{1-\left(\frac{L}{2b}\right)^2} \tag 4$$ Condition to have BC=$L$ : $\quad (x_1-x_2)^2+(y_1-y_2)^2=L^2$ $$\left(a\frac{a^2-\sqrt{b^4+(a^2-b^2)L^2}}{a^2-b^2}-a\sqrt{1-\left(\frac{L}{2b}\right)^2}\right)^2+ +\left(b\sqrt{1-\left(\frac{a^2-\sqrt{b^4+(a^2-b^2)L^2}}{a^2-b^2}\right)^2}-a\sqrt{1-\left(\frac{L}{2b}\right)^2}\right)^2=L^2$$ this equation contains only the unknown $L$.

It can be simplified by normalization, with $x=\frac{L}{a}$ , $\beta=\frac{b}{a}$. This leads to an equation with parameter $\beta$ and unknown $x$. But the benefit is not large because transforming the equation with radicals leads to a polynomial equation of degree higher than 5.

Thus, there is no closed form for the unknown $L$. Nevertheless, one can easily solve it for $L$ with numerical calculus. Then the coordinates of the points are computed with Eqs.$(2-4)$. An example is shown on the figure below.