Proving a property of an ellipse and a tangent line of the ellipse 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.
 A: 
Required to prove that the acute angles made by each foci to the tangent are equal
A: 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.  
A: 
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 $(\frac \pi 2, \pi)$.
