Proving a property of an ellipse and a tangent line of the ellipse

Question

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.

Answer 1

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.

Answer 2

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 $(\frac \pi 2, \pi)$.