TITLE: Conjecture: in an ellipse with major axis AB, the projection I of one of the foci on any tangent is such that AIB is right angle
QUESTION [19 upvotes]: I think I have found out some property of an ellipse. Define the following:

The point $P$ belongs to the ellipse.
The tangent line $\ell$ goes through $P$.
The point $F$ is one of the foci.
The line $m$, through $F$, is perpendicular to the line $\ell$ at point $I$.
$A$ and $B$ are the vertices of the ellipse (that is, the endpoints of the major axis).


In this case,
The $\angle AIB$ is always $\pi/2$. (I think.....)

But I didn't prove yet...

Is it true?
Could anybody give me some advice?

Thanks in advance.
Edit1 For jmerry

Edit2 For jmerry
Really thanks, makes me fun!

REPLY [4 votes]: Not a strict proof, an interesting observation though.
There is a method to draw an ellipse by folding paper. See this video: Folding a Circle into an Ellipse. In case the link dies, of if you want more information, search for folding paper ellipse or so.
The method goes like this:

Draw a circle on a piece of paper, $G$ is the center.
Choose a point $F$ inside the circle.
Repeat many times ($n=0,1,2,3,…$) the following:


choose a point $J_n$ on the circle;
fold the paper so $J_n$ meets $F$;
unfold, denote the folding line $l_n$.

An ellipse will appear with $F$ and $G$ as foci; each $l_n$ will be some tangent line.

In the question you're working in reverse! You have the ellipse, you have $F$ and $G$. You choose some $l$ and find corresponding $J$. All possible $J$-like points should then form a circle which I denote $\mathcal{J}$; we expect $G$ to be its center.
How about $I$ and all possible $I$-like points (the set I denote $\mathcal{I}$)? From the fact that $\overrightarrow{FI} = \frac 1 2 \overrightarrow{FJ}$ we conclude $\mathcal{I}$ should be a homogeneous dilation of $\mathcal{J}$ with the center $F$ and the ratio $\frac 1 2$. So all possible $I$-like points also form a circle. Applying the dilation to the center of the circle $\mathcal{J}$ (i.e. to the point $G$) we obtain the center of the circle $\mathcal{I}$. In your case it's the origin $O$.
Now consider tangent lines at $A$ and at $B$. It's easy to see these two points must belong to $\mathcal{I}$. Knowing $O$ is the center of $\mathcal{I}$, we conclude $AB$ is the diameter.
Your $I$ is on the circle $\mathcal{I}$, the diameter is $AB$. Therefore $\angle AIB = \frac \pi 2$.