Is there a theorem of intersecting chords in an ellipse? 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 $f$.
 A: 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.


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.

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.
A: Given a chord $AB$ and a point $P$ on it, then consider the diameter $A'B'$ of the ellipse parallel to $AB$. Let $r=AP/A'B'$, $s=BP/A'B'$, then $r\times s$ is constant (independent of the chord direction).
The proof is that an ellipse is a scaled version of a circle. Scalings do not preserve distances but they do preserve the ratio of lengths of parallel lines. The original theorem about circles can be recast to state that the product of the ratios of chord segments with the radius (or diameter) is constant. Hence this remains true for the ellipse.
Corollary: If two chords $AB$, $CD$, of an ellipse meet at a fixed point $P$, then $$\frac{AP.PB}{CP.PD}=\frac{(A'B')^2}{(C'D')^2}.$$
A: 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.
