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Ellipse

The angle between tangent line l and F2P is the same as between line l and PF1 (down in the picture). Thats why a light beam from one focal point always gets to the other ofcal point. But what if a light beam in this ellipse sets off from a point elsewhere than the focal points, lets say just a bit right from F2. Is the angle of incidence still the same as of refraction? If so, why, because I dont see how I could prove this.

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  • $\begingroup$ As a mathematician, you don't need to prove this. This is the assumption in this problem. With that assumption, you can prove that the "light beam" from one focus ends up in the other focus. There may well be interesting theorems about light beams going elsewhere and bouncing (with the same assumption!) and satisfying some exciting conditions. On the other hand, as a physicist, you can question the validity of this assumption in real life, but that is where this stops being a mathematical problem. $\endgroup$ – Stinking Bishop Feb 3 at 19:44
  • $\begingroup$ BTW, it is (roughly) true: the light does bounce, even with curved mirrors, so that the angle of incidence is the same as the angle of reflection. Or at least it does in the first approximation. $\endgroup$ – Stinking Bishop Feb 3 at 19:46
  • $\begingroup$ youtube.com/watch?v=4KHCuXN2F3I&app=desktop $\endgroup$ – user645636 Feb 4 at 13:26
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The angle of incidence is the same as the angle of reflection, but this is a law of geometrical optics which has nothing to do with mathematics. The above angles are measured between each ray of light and the normal line at the reflection point.

The mathematical part, for an ellipse, lies in the fact that the normal line at a point $P$ of the ellipse is the bisector of $\angle F_1PF_2$.

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