By user143993

2019-02-05 06:58:53 8 Comments

enter image description here

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...

  1. Is it true?
  2. Could anybody give me some advice?

Thanks in advance.

Edit1 For jmerry

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Edit2 For jmerry

Really thanks, makes me fun!

enter image description here


@Kamil Maciorowski 2019-02-05 21:47:44

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:

  1. Draw a circle on a piece of paper, $G$ is the center.
  2. Choose a point $F$ inside the circle.
  3. Repeat many times ($n=0,1,2,3,…$) the following:
    1. choose a point $J_n$ on the circle;
    2. fold the paper so $J_n$ meets $F$;
    3. unfold, denote the folding line $l_n$.
  4. 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 $\angle AIB = \frac \pi 2$.

@jmerry 2019-02-05 07:16:13

Yes, it's true.

Some advice: the locus of all points $I$ such that $\angle AIB$ is a right angle is the circle with diameter $AB$. So, then, if you can show that $OI=OA=OB$ where $O$ is the center of the ellipse, you'll have it.

Next, add some more things to your diagram. The other focus (let's call it $G$) looks like a good place to start - after all, that lets us use nice facts like the segments from $P$ to the two focuses making the same angle to the tangent line $\ell$, and that $PF+PG=AB$.

So now, how can you use those facts? Don't be afraid to extend lines and find new intersections.

[Added in edit]
All right, you've drawn some more things in, and some equal angles. Since we're looking at $I$ in particular, how about naming that point where $GP$ and $FI$ intersect? You've already labeled an angle there, so naming the point will make it easier to work with.

(I called it $J$ in my diagram)

Oh, and since we're interested in the length of $OI$, that's another segment we should draw.

@user143993 2019-02-05 07:34:06

I would like to listen more about "Next ~ intersections." Could you please check my edited question?

@user143993 2019-02-05 07:36:01

I have just added another diagram.

@jmerry 2019-02-05 09:01:22

Huh - never heard that name before. Of course, the theorem I was thinking of, attributed to Thales, is far older than Monge.

@Jean Marie 2019-02-06 21:22:09

The circle you mention in your first sentence is not Monge's circle as I thought at first but the "auxiliary circle" (I have found back its name just now) :

@Thomas 2019-02-05 07:26:08

Hint: The only way for the angle $AIB$ to always be $\frac{\pi}{2}$ is if the point $I$ lies along a circle with radius $|B|$ (Check out Thales's Theorem). In other words, the distance from I to the origin must be constant.

@user143993 2019-02-05 07:42:58

Thanks a lot. Can I have more hint for proving OI=OA, where O is the origin

@Thomas 2019-02-05 07:52:15

If you have an expression for the $x$ coordinate of I and the $y$ coordinate of $I$, then showing $\sqrt{x^2 + y^2} = Constant = |A|$ is sufficient. I'm sure there is a much more elegant way to do it geometrically but one way is to express the top half of the ellipse as a function and then calculate a general tangent line to the ellipse (using its derivative) and find a line normal to the tangent (by reciprocating the slope of the tangent line) that goes through the focus. Then find where the tangent and the normal intersect, that is the coordinates for $I$.

@Thomas 2019-02-05 07:52:21

Note: if you have proven it for the top half, it must be true for the bottom half as well (by symmetry)

@user143993 2019-02-05 08:18:15

Thanks for helping! Really thanks.

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