Is this ratio in an ellipse constant? Can it be proven synthetically?

Question

Is the following assertion true? How can it be proved? I'm interested in any proofs, but in particular interested if it can be proven synthetically (i.e. without coordinates).

Call the endpoints of the major axis $A$ and $A'$. For any point $P$ on the ellipse, let drop the perpendicular from $P$ to the major axis at $Q$. Then $$\frac{|PQ|^2}{|AQ||A'Q|}$$ is a constant (and indeed equal to the square of the ratio of the major axis to the minor axis).

Background: This fact is used in a synthetic proof that Archimedes' Trammel draws an ellipse. Archimedes proof is lost, but, if we can prove this fact, ideally using synthetic (Euclidean) geometry, we can reconstruct a proof.

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For background, an analytic proof is simple. Given an ellipse with center at the origin $\frac {x^2}{a^2} + \frac{y^2}{b^2} = 1$, the claim is that $$x^2 = k \cdot (b-y)(b+y)$$ which follows immediately by setting $k := \frac{a^2}{b^2}$.

Answer 1

Let $a$ and $b$ be the major and minor semi-axis lengths of the ellipse. Construct a circle of radius $a$ about the center of the ellipse, so that $AA'$ is a diameter of the circle. Extend $QP$ to intersect the circle at $P'.$

Then $$\frac{\lvert PQ\rvert}{\lvert P'Q\rvert} = \frac ba$$ and $$\lvert P'Q\rvert^2 = \lvert AQ\rvert\lvert A'Q\rvert$$

from which it follows that

$$\frac{\lvert PQ\rvert^2}{\lvert AQ\rvert\lvert A'Q\rvert} = \frac{b^2}{a^2}.$$

Answer 2

An ellipse $ABP$ can be defined as the intersection between a right circular cone and a plane. The plane $VAB$ through the axis of the cone and perpendicular to the plane of the ellipse intersects the ellipse at major axis $AB=2a$ (see figure below).

Consider then a plane through $P$ parallel to the base of the cone (and thus perpendicular to plane $VAB$), intersecting the cone at circle $A'B'P$, where $AA'$ and $BB'$ are two generatrices of the cone, and intersecting $AB$ at $H$. Line $PH$ is perpendicular to plane $VAB$ and is thus perpendicular to both $AB$ and $A'B'$.

By the intersecting chords theorem we know that $PH^2=A'H\cdot B'H$. But, on the other hand, we have by similitude: $$ A'H:AH=BC:AB, \quad\hbox{that is:}\quad A'H={BC\over AB}\cdot AH; $$ and: $$ B'H:BH=AD:AB, \quad\hbox{that is:}\quad B'H={AD\over AB}\cdot BH. $$

Plugging these into the previous formula we obtain:

$$ \tag1 PH^2={AD\cdot BC\over 4a^2}\,AH\cdot BH. $$

In particular, if we choose plane $A'B'P$ midway between $A$ and $B$, so that $H$ is the midpoint of $AB$ and $AH=BH=a$, while $PH=b$ is the minor axis of the ellipse, we get:

$$ b^2={AD\cdot BC\over 4a^2}\,a^2, \quad\implies\quad AD\cdot BC=4b^2. $$

We can then rewrite $(1)$ as:

$$ \tag2 PH^2={b^2\over a^2}\,AH\cdot BH. $$

Answer 3

In circle $c_1$ we have:

$IQ^2=AQ\cdot A'Q\space\space\space\space\space\space(1)$

Triangles SIQ and SRO are similar,Also $\triangle SPQ\sim\triangle SFO$ so we have:

$\frac{PQ}{FO}=\frac{SQ}{SO}$

$\frac{IQ}{RO}=\frac{SQ}{SO}$

$\Rightarrow \frac {PQ}{FO}=\frac {IQ}{RO}\space\space\space\space\space\space(2)$

$\frac {PQ}{IQ}=\frac{FO}{RO}\space\space\space\space\space\space(3)$

$RO=a$

$FO=b$

$\Rightarrow \frac{PQ^2}{b^2}=\frac{IQ^2}{a^2}\space\space\space\space\space\space(4)$

$\Rightarrow PQ^2=\frac{IQ^2}{a^2}\cdot b^2\space\space\space\space\space\space(5)$

From (1) and (5) we get:

$\frac {PQ^2}{AQ\cdot A'Q}=\frac{b^2}{a^2}$