Is there a theorem of intersecting chords in an ellipse?

Question

I found a well known theorem that if $A,B, C$ and $D$ are on the circumference of a circle and $AB\cap CD=P$ then $AP\cdot BP=CP\cdot DP$ . Is there anything generalization of it to an ellipse? Maybe something that in a given ellipse, if $P$ divides two line segments to parts of length $a,b,c,d$ and major axis has length $e$ and minor axis has length $f$ then there is some algebraic identity that connects $a,b,c,d,e$, and $f$.

Answer 1

As I know you know (based on the original draft of your question), the "best" way to think of the Intersecting Chords Theorem is as an aspect of the Power of a Point relative to a circle:

If a line through point $P$ meets a circle at points $M$ and $N$, then the product of (signed) distances from $P$ to those points is a constant (called the power of $P$) that depends only upon the $P$'s position relative to the circle. Specifically, if $r$ is the radius of the circle, then $$|\overrightarrow{PM}||\overrightarrow{PN}| \;=\; |\overline{OP}|^2 - r^2 \tag{$\star$}$$

(The notation $|\overrightarrow{PM}|$ is meant to indicate "signed length": $|\overrightarrow{PM}|$ and $\overrightarrow{PN}|$ have the same sign (respectively, opposite signs) if vectors $\overrightarrow{PM}$ and $\overrightarrow{PN}$ point in the same direction (respectively, opposite directions).)

The Power of a Point concept turns out to be pretty useful (dare I say, "powerful") in geometry, and is something of a gateway result that helps motivate things like the Poincaré model of hyperbolic geometry. (But I digress.) So far as I know, there's no direct analogue of $(\star)$ available for general conic sections. By this I mean: I don't know of a function that converts (signed) lengths of arbitrary segments from $P$ into a "power"-like constant. That doesn't mean that there isn't anything to be said about this kind of thing. For instance, consider when the point $P$ is a focus (denoted $O$ below) in an arbitrary conic.

If a line through focus $O$ meets a conic at points $M$ and $N$, then, defining $m := |\overrightarrow{OM}|$ and $n := |\overrightarrow{ON}|$, and with $r$ the length of the conic's semi-latus rectum, $$r^2 ( m - n )^2 = 4 m^2 n^2 \tag{$\star\star$}$$ Alternatively, introducing the notation $\overline{x} := 1/x$ for reciprocation (think of it as a fraction with an understood "$1$" in the numerator), $$(\overline{m} - \overline{n})^2 = 4\;\overline{r}^2 \tag{$\overline{\star\star}$}$$

Proof. The polar-coordinate equation of a conic of eccentricity $e$ with its focus at the origin (and corresponding vertex in the right half-plane) is: $$\rho = \frac{r}{1+e\cos\theta}$$ So, with $\phi$ the angle that line $\overleftrightarrow{MN}$ makes with the axis of the conic, we have (with an appropriate tweak in sign for $n$) $$\begin{array}{lcl} m := \dfrac{r}{1+e\cos\phi}&\quad\to\quad& \overline{m} = \overline{r}\;(\phantom{-}1+e\cos\phi)\\[4pt] n := \dfrac{-r}{1+e\cos(\phi+\pi)} = \dfrac{-r}{1-e\cos\phi} &\quad\to\quad& \overline{n} = \overline{r}\;(-1+e\cos\phi) \end{array}$$ and the result follows. (We don't write simply $\overline{m}-\overline{n}=2\overline{r}$, since $r$ (and $\overline{r}$) should be considered unambiguously positive.) $\square$

Thus, associated with the focus of a there is a constant value of a simple arithmetic combination of signed segment lengths through that focus. Unfortunately, that same arithmetic combination does not generate associated constants for non-focus points. So, there's no "power" here. Nevertheless, there's more to say ... it's just complicated to say.

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In this figure, we consider lines $\overleftrightarrow{MN}$ through an arbitrary point $P$, which we can take to lie on a similar ellipse with common eccentricity $e$ and common focus $O$, and with semi-latus rectum $s$. Of all the lines through $P$, one is a focal chord through $O$ (making angle $\theta$ with the conic's axis) that we'll denote $\overline{FG}$, and we define $$\begin{align} c &:= \cos\theta \\[4pt] f &:=|\overrightarrow{PF}| = |\overrightarrow{OF}| - |\overrightarrow{OP}| = \frac{r}{1+ce} - \frac{s}{1+ce} = \frac{r-s}{1+ce} \\[4pt] g &:=|\overrightarrow{PG}| = |\overrightarrow{OG}| -|\overrightarrow{OP}| = \frac{-r}{1-ce}-\frac{s}{1+ce} = - \frac{r (1+ce) + s(1-ce)}{1 - c^2 e^2} \end{align}$$

We can generalize $(\overline{\star\star})$ to a fourth-degree polynomial equation with coefficients determined by $f$, $g$, $c$, and $e$. As it turns out, though, some auxiliary values greatly help reduce clutter in the presentation of that polynomial; we make these seemingly-unmotivated definitions ...

$$ u := \phantom{-}\frac{\overline{f}}{1 + c e} + \frac{\overline{g}}{1 - c e} \qquad\qquad v := \frac{\overline{f}}{1+c e} - \frac{\overline{g}}{1 - c e} $$

... so that we can give the polynomial equation in this form:

$$\begin{array}{c} \left(\;\; \begin{array}{c} e^2 (u^2 - v^2) \left(\;(\overline{m} + \overline{n})^2 - u^2 (1 - c^2) \;\right)\\ + \left(\;4 \overline{m} \overline{n} - (u^2 - v^2)\;\right) \left(\; (c u + e v)^2 - u^2 (1-c^2)\;\right) \\ \end{array} \;\;\right)^2 \\[8pt] + \; 4 p^2 \; (1-c^2) \; (c u + e v)^2 \; \left(\;4 \overline{m} \overline{n} - (u^2 - v^2)\;\right)\; \left(\;4 \overline{m} \overline{n} - (u^2 - v^2) (1 - e^2)\;\right) \quad = \quad 0 \end{array}$$

(Note that we can replace occurrences of $1-c^2$ with $\sin^2\theta$.)

• For a circle (ie, $e = 0$), the equation reduces to $\overline{m}\overline{n}(s^2-r^2) = 1$, which is equivalent to $(\star)$.

• For $P$ coincident with focus $O$ (ie, $s=0$, so that $u = 0$ and $v = 2\overline{r}$), we get $(\overline{\star\star})$ back.

• I haven't found many other particularly-compelling general cases that lead to such dramatic reductions.

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Proof. While there may be more-direct approaches, I did the following ...

The Cartesian form of the equation for our focus-at-origin ellipse is $$x^2 (1- e^2) + y^2 + 2 e r x - r^2 = 0 \tag{1}$$

Moreover, we have $$P = \frac{s}{1+e\cos\theta}(\cos\theta, \sin\theta) \qquad M = P + m (\cos\phi, \sin\phi) \qquad N = P + n (\cos\phi,\sin\phi)$$ where $m$ and $n$ are roots of the quadratic equation obtained by substituting $P+z(\cos\phi,\sin\phi)$ into $(1)$:

$$z^2 (1 + c e) (1 - e^2\cos^2\phi) \;+\; 2 z ( \cos\theta( e r (1+ c e) + c s (1- c e)) + s \sin\theta \sin\phi) - (r - s) (r (1+c e) + s(1 - c e)) = 0 \tag{2}$$

We could solve for $z$ explicitly, but we won't bother; instead, we observe (by Vieta's formulas) that

$$\begin{align} m + n &= - \frac{2 ( \cos\phi ( e r (1+ c e) + c s (1- c e)) + s \sin\theta \sin\phi)}{(1 + c e) (1 - e^2\cos^2\phi)} \tag{3a}\\[6pt] m n &= \frac{(r - s) (r (1+ce) + s(1 - c e))}{(1 + c e)(1 - e^2\cos^2\phi)} = fg\;\frac{1-e^2\cos^2\theta}{1 - e^2\cos^2\phi} \tag{3b} \end{align}$$

The fact that, for $e=0$, equation $(3b)$ becomes $mn = fg$ is essentially the Power of a Point theorem for circles; but, in general, we see that the sums and products of $m$ and $n$ are non-constant values that depend upon the parameter $\theta$ that governs the direction of $\overleftrightarrow{MN}$ through $P$. However, since $(3a)$ and $(3b)$ constitute two equations involving the parameter $\theta$, we can eliminate that parameter. I used little trig to rewrite the $(3a)$ as a polynomial in $\cos\theta$, and then the method of resultants to eliminate $\cos\theta$. It took a bit of massaging to achieve the form presented above. (In Mathematica, the fully-expanded raw result was an expression in approximately $1300$ terms.) I believe that more massaging can provide an even-better form ---ideally, one that reveals some geometric meaning--- but I'll leave it as-is for now.

Answer 2

Two constants $ \alpha, \beta $ are needed in the ellipse 'generalization':

$$ \dfrac{ AP \cdot PB}{ CP \cdot PD } = {\dfrac {\cos ^2\alpha}{\cos^2 \beta}} $$

when the segments are projected on a plane of a $ circle $ as $ ap,pb, cp,pd $ cutting at $p$ making inclination angles $ \alpha, \beta $ to the plane containing the ellipse.

So if the generating inclinations are related as:

$$ \dfrac {\cos \alpha}{\cos \beta} = c, $$

then the product of segments can be constant.