analytic geometryeuclidean-geometry
I know that if a line $r$ is tangent to an ellipse with foci $F$ and $F'$ at some point $T$, then $r$ is perpendicular to the bisector of the angle $FTF'$.
Is there a simple proof of that? By 'simple' I mean a proof suitable for high school students.
Let the foci be $F_1$ and $F_2$ and let a point of intersection be $P$. The tangent to the ellipse is the external angle bisector of $\angle F_1PF_2$, and the tangent to the hyperbola is the internal angle bisector.
Proving these statements without calculus is onerous. Here's a sketch of a partial proof for the statement about ellipses, to give the main ideas:
Let $F_1,F_2,P$ be as above. Let $b$ be the external bisector of $\angle F_1PF_2$; we want to show that $b$ is tangent to the ellipse. By the definition you've been given, that means we want to show that $b$ intersects the ellipse only at $P$. Consider first a point $Q$ on the other side of $b$ from the foci; I claim that $Q$ is not on the ellipse. To show this, draw $F_2Q$, meeting $b$ at $R$; also reflect $F_2$ in $b$ to obtain its image $F_2'$. Note that $F_1PF_2'$ are collinear because $b$ is the external angle bisector. Then $$ F_1Q+QF_2 = F_1Q + QR + RF_2 = F_1Q + QR + RF_2' > F_1F_2' = F_1P + PF_2' = F_1P + PF_2 $$ So $Q$ is not on the ellipse. I'm skipping some details here, notably why it has to be $>$ and not $\ge$ in the triangle inequality here. After you fill in that detail, and if you believe that the ellipse is a continuous curve, then we know that the ellipse doesn't cross $b$, so it is tangent to the ellipse. Then give a similar analysis for hyperbolas.
(We have not proved that conic sections have only one tangent at each point; I think I proved that once from your definition, but I only remember that it was unpleasant... and maybe my proof was wrong anyway.)
The nice way to prove these statements is with the notion of gradient from multivariable calculus: the direction of the gradient of the scalar field "distance from $F_i$" is obviously a unit vector pointing away from $F_i$; from this and the definition of an ellipse as a level curve of the sum of two such scalar fields it's clear that the normal to the ellipse bisects the desired angle. (On preview, along the lines of achille hui's answer.)
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
The unique point at which the tangent line hits the ellipse must be the point on that line at which the sum of the distances to the foci is minimized. Given two points on one side of a line, to minimize the sum of the distances in this way to any point on the line the best thing is to reflect one point over the line, and then the shortest path between them is a straight line. This shows the reflection property of ellipses, and what you are looking for is immediate from that.
Hope that helps.
Greg