Constant ratios of parallel elliptical sections of an ellipsoid

Question

QUESTION

Given an ellipsoid of semi-axis $a, b, c$, for any $z $$\in$ $(-c, c)$ we get an elliptic section with semi-axis $a_z, b_z$ and parametric equation

$$ \begin{split} x &= a_z\cos(t)\\ y &= b_z\sin(t) \end{split} \quad \text{for} \quad 0 \leq t \leq 2\pi $$

Let us consider only the first quadrant of such ellipses, i.e. $ 0 \leq t \leq \dfrac{\pi}{2}$ and define the ratio

$$ q_z(t) = \frac{\Delta p_z(t)}{P_z/4} $$

where $\Delta p_z(t)$ is the arclength of the ellipse as a function of the angle spanned and $P_z$ is the ellipse's perimeter.

Can the following statement be proven?

$$q_{z_{1}}(t) = q_{z_{2}}(t) \quad \text{for}\quad z_1,z_2,t \quad \text{s.t.} \begin{cases} z_1 \neq z_2\\ z_1, z_2 \in (-c, c)\\ 0 \leq t \leq \frac{\pi}{2} \end{cases}$$

To put it another way: given an angle $t$ and two parallel elliptical sections of the same ellipsoid, is the ratio between the arclength of $t$ and the ellipse's perimeter the same on both ellipses?

BACKGROUND

Never worked with elliptic integrals before. My 'intuitive' approach is the following:

1. Prove that if $r_{z_{1}} = r_{z_{2}}$ then $q_{z_{1}}(t) = q_{z_{2}}(t)$ (where $r_z=\frac{a_z}{b_z}$)
2. Prove that $r_z$ is constant throughout $z$

But I have no clue on how to approach step 1.

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EDIT

Found my proof of step 2 of the 'intuitive' approach.

The ellipsoid equation can be written as

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$

The intersection of plane $y=0$ with the ellipsoid's equation gives us a relation between $z$ and the semi-axis $a_z$ of the elliptical section cut by the plane $z$

$$ \frac{a^2_z}{a^2} + \frac{z^2}{c^2} = 1 \xrightarrow{} a_z = a\sqrt{1 - \frac{z^2}{c^2}} $$

and the same applies for $x=0$ and semi-axis $b_z$

$$b_z = b\sqrt{1 - \frac{z^2}{c^2}}$$

proving then that $r_z = \frac{a_z}{b_z} = \frac{a}{b}$ which is constant throughout $z$ for any given ellipsoid.

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EDIT 2

Having seen that $r_z$ is constant means that all the parallel elliptic sections of an ellipsoid are (as expected) proportional to each other. If we can prove that this proportional scaling of the ellipses also holds for the perimeter and the arclength then we can also prove the statement of the question.

Answer 1

The ellipsoid has the implicit equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$

and a $z$ section,

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1-\frac{z^2}{c^2}.$$

So the axis of the sections are $\dfrac{ac}{\sqrt{c^2-z^2}}$ and $\dfrac{bc}{\sqrt{c^2-z^2}}$ and all sections are homothetic.

This answers your question.

Answer 2

In the end it came down to proving that if two ellipses $\psi_1, \psi_2$ have the same eccentricity then $$\frac{arclength_{\psi_1}(\theta)}{perimeter_{\psi_1}} = \frac{arclength_{\psi_2}(\theta)}{perimeter_{\psi_2}}$$ Interestingly, if two ellipses share the same eccentricity, it can be proven that

$$\frac{perimeter_{\psi_{1}}}{perimeter_{\psi{2}}} = \frac{a_1}{a_2} = \frac{b_1}{b_2} = \left\{ \text{by proof below}\right\} = \frac{arclength_{\psi_{1}}(\theta)}{arclength_{\psi_{2}}(\theta)}$$

which is some kind of Thales theorem for arc sections of ellipses.

The proof of the first equalities above comes from the expansion series of an ellipse's perimeter

$$p = 2a\pi \left( \sum_{n=1}^{\infty}\frac{(2n)!^2}{(2^n\cdot n!)^4}\cdot\frac{e^{2n}}{2n - 1}\right)$$

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STATEMENT 1

Given the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$ and two parallel planes $$\begin{cases}z &= z_1 \\ z&= z_2\end{cases} \quad \text{where} \quad z_1,z_2 \in (-c, c)$$ let us consider the conic sections (ellipses in this case) produced by the intersection of said planes and the ellipsoid and represent them as $$\begin{cases}\frac{x^2}{a_{z_{1}}^2} + \frac{y^2}{b_{z_{1}}^2} &= 1 \\ \frac{x^2}{a_{z_{2}}^2} + \frac{y^2}{b_{z_{2}}^2} &= 1\end{cases}$$ Given the ratio $r_z = \frac{a_z}{b_z}$, prove that $$r_{z_{1}} = r_{z_{2}} \quad \forall z_1,z_2 \in (-c, c)$$

PROOF

Refer to the first edit of the question.

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STATEMENT 2

For the ellipsoid of Statement 1, consider the ratio $$q_z(\theta) = \frac{\Delta p_z(\theta)}{P_z/4}$$ where $\Delta p_z(\theta)$ is the arclength of the elliptic section (given by the intersection of the ellipsoid and the plane z) as a function of the 'central angle' $\theta$ and $P_z$ is the perimeter of said ellipse.

Given the ratio $r_z$, prove that $$r_{z_{1}} = r_{z_{2}} \implies q_{z_{1}}(t) = q_{z_{2}}(t) \quad \forall z_1,z_2 \in (-c, c)$$

PROOF

An ellipse can be described by the parametric equation

$$\psi (t) = (x(t), y(t)) = (a\cos(t), b\sin(t)) \quad \text{for} \quad 0\leq t\leq 2\pi $$

where $t$ is related to the 'central angle' $\theta$ as follows

$$\tan(\theta) = \frac{y}{x} = \frac{b}{a} \tan(t)$$

The arclength of an ellipse expressed in terms of parameter $t$ can be expressed as

$$s(t) = a\cdot E\left[ t, \sqrt{1 - \frac{b^2}{a^2}}\right] = b\cdot E\left[ t, \sqrt{1 - \frac{a^2}{b^2}}\right]$$

where $E[z, m]$ is the incomplete elliptical integral of the second kind. Now, the ratio $q_z(t)$ of the question can be expressed as

$$q_z(t) = \frac{4\cdot s_z(t)}{s_z(2\pi)} = 4\cdot \frac{a_z\cdot E\left[ t, \sqrt{1 - \frac{b_z^2}{a_z^2}}\right]}{b_z\cdot E\left[ 2\pi, \sqrt{1 - \frac{a_z^2}{b_2^2}}\right]}$$

the substitution $r_z = \frac{a_z}{b_z}$ leaves

$$q_z(t) = 4r_z\cdot \frac{E\left[ t, \sqrt{1 - \frac{1}{r_z^2}}\right]}{E\left[ 2\pi, \sqrt{1 - r_z^2}\right]}$$

which proofs that

$$ q_{z_{1}}(t) = q_{z_{2}}(t) \quad \text{if} \quad r_{z_{1}} = r_{z{2}}$$