How to prove geometrically that if we have a tangent of ellipse with focus F and F' in point P, that tangent is bisector of the angle between a line joining focus F to point P and the line F'P outside the ellipse?
Geometry – How to Geometrically Prove the Focal Property of Ellipse
conic sectionsgeometry
Related Solutions
Required to prove that the acute angles made by each foci to the tangent are equal
There is a simple way to construct a second tangent, perpendicular to the first one you constructed at $P$. Let the normal at $P$ intersect again the ellipse at $N$ and let $M$ be the midpoint of $PN$. The line passing through $M$ and the center $O$ of the ellipse will then intersect the ellipse at two points $Q$ and $Q'$, with the property that the tangents at both points are parallel to $PN$, and hence perpendicular to the first tangent.
As I wrote in a comment, the intersections between the tangents at $Q$ and $Q'$ and the tangent at $P$ lie on the director circle of the ellipse.
Best Answer
Suppose that you would like to find the shortest way from $A$ to $B$ that touches the angled line. Observe that if we were to reflect $B$, then the distance from $A$ to $B$ and from $A$ to $B'$ are the same, but the shortest path from $A$ to $B'$ is a straight line, the solution follows. The interesting part is that, because of the reflection, the angles marked red are equal.
Let $f(X) = |FX| + |F'X|$ so that the ellipse is exactly the set $f^{-1}\big(f(P)\big)$. Note that for any point $X$ outside of the ellipse we have $f(X) > f(P)$, on the ellipse we have equality, and inside $f(X) < f(P)$. Moreover, for any point $Y$ on the tangent setting $Y=P$ minimizes $f(Y)$, the sum of distances from $F$ and $F'$, exactly as in the first diagram. Hence, the marked angles are again equal.
I hope this helps $\ddot\smile$