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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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    @W12: Other than the algebraic definition of an ellipse, you can define it geometrically. For example, it is the locus of the points whose distance from two fixed points (the foci, here $F$ and $F'$) add up to a constant. You can also consider it a s a special case of conic sections2012-10-15

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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 $\frac{b}{ae\sin(\theta)}$.2012-10-15
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/Ellipse with a tangen

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

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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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    @MhenniBenghorbal: My pleasure.2012-10-15