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Suppose that there is line $l$ that is tangent to an ellipse $A$ at point $\,P\,$.

The ellipse has the foci $F'$ and $F$.

One then creates two lines - each from each focus to the tangency point $\,P\,$ .

What I want to prove is that the acute degree formed at $P$ between $l$ and the line segment $F'P$ equals the acute degree formed between $l$ and the line segment $FP$ .

How would I be able to prove this?

(ellipse has a horizontal axis as a major axis.)

Edit: line $l$ and the corresponding $\,P\,$ can be set arbitrarily (they just need to meet the aforementioned condition), so what I want to prove is for all possible cases.

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So you do not want "two lines that intersect at one point P on the ellipse: you want P to be the tangency point of the tangent $\,l\,$ and the ellipse...don't you? – DonAntonio Oct 15 '12 at 3:29
@DonAntonio Edited my question. Can you have a look over it? – W12 Oct 15 '12 at 3:31
I edited your question trying to correct the language. Check if this fits your intention. Remember that you can always try to write down your question also in your native language. Perhaps someone here understands it and can help. – DonAntonio Oct 15 '12 at 3:41
@W12: Since rotation preserves angles, you can assume the ellipse is axis-aligned and then you can prove it analytically. I'm sure there are elegant pure geometric proofs, but they depend on how you define an ellipse and what properties of an ellipse are assumed to be known. – S.B. Oct 15 '12 at 3:51
@S.B. What do you mean by "how you define an ellipse..?" Isn't ellipse the same ellipse we know of? I am assuming the general use of ellipse.... – W12 Oct 15 '12 at 4:04
up vote 2 down vote accepted

Follow these steps:

1) Find the equation of the tangent at the point $P$.

2) Find the direction vector $v$ of the tangent line.

3) Construct the vector $u = P- F' $.

4) Construct the vector $ w = P - F $.

5) Find the angle $\theta_1$ between the vectors $v$ and $u$.

6) Find the angle $\theta_2$ between the vectors $v$ and $w$.

7) Compare the two angles.

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How can I find the equation of the tangent at the point P when line $l$ and $P$ can be set arbitrarily? – W12 Oct 15 '12 at 4:08
define $l:y=mx+n$ and point $P(x_0,y_0)$ – Salech Alhasov Oct 15 '12 at 4:22
@SalechAlhasov: Thanks for answering him. – Mhenni Benghorbal Oct 15 '12 at 4:34
@MhenniBenghorbal: My pleasure. – Salech Alhasov Oct 15 '12 at 4:37

/Ellipse with a tangen

Required to prove that the acute angles made by each foci to the tangent are equal

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enter image description here

Without any loss of generality, we can assume the ellipse to be $\frac {x^2}{a^2}+\frac{y^2}{b^2}=1$ and P$(a\cos \theta, b\sin \theta)$, F$(ae,0)$ and F'$(-ae,0)$.

Applying derivative wrt $x, 2x+\frac{2ydy}{dx}=0$

So, the gradient $m_1$ at P$(a\cos \theta, b\sin \theta)$ is $$-\frac{a\cos \theta}{ b\sin \theta}$$

The gradient $m_2$ of $FP$ is $$\frac{b\sin \theta-0}{a\cos \theta-ae}$$

So, if the angle between $l$(tangent) and FP is $\phi$

then $$\tan \phi=\pm \frac{m_1-m_2}{1+m_1m_2}=\pm \frac{b^2\sin^2\theta+a^2\cos^2\theta-a^2e\cos\theta}{-abe\sin \theta}$$

Similarly for F'P and the tangent$(l)$,

$$\tan \phi'=\pm \frac{b^2\sin^2\theta+a^2\cos^2\theta-a^2e\cos\theta}{abe\sin \theta}$$

If $$z=\frac{b^2\sin^2\theta+a^2\cos^2\theta-a^2e\cos\theta}{abe\sin \theta},$$

if $z\ge 0,$ the acute angle is either cases will be $\tan^{-1}z$ taking the principal value i.e in $[0,\frac \pi 2]$

else the acute angle is either cases will be $\tan^{-1}(-z)=\pi-\tan^{-1}z$, the value of $\tan^{-1}z$ lies in $(\frac \pi 2, \pi)$.

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Nice. I was about to type up the exact same approach but then saw you beat me to it. In my calculations, however, I had the slope of the tangent at $P$ as being $-\frac{b\cos(\theta)}{a\sin(\theta)}$. You need to keep the $a$ and $b$ terms in your derivative and they would cancel out in this way. Your resulting equations for $\tan(\phi)$ and $\tan(\phi^\prime)$ would then reduce to $\frac{b}{ae\sin(\theta)}$. – cheepychappy Oct 15 '12 at 5:43

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