Tangent to conic stays outside or between (elementary) 
Let $f$ be a function from $\mathbb{R}^2$ to $\mathbb{R}$ such that $f(X) = |XA| + |XB|$, where $A,B \in \mathbb{R}^2$. Let $T$ be a point on the ellipse $\varepsilon\colon f(X)=k$, for some $k\ge |AB|$. If $C \neq T$ is a point on the tangent line to $\varepsilon$ through $T$, then $f(C) > f(T)$. 

The same goes for $g(X) = \bigl||XA| - |XB|\bigr|$, changing ellipse with hyperbola. 
I've seen a proof to this (or a similar theorem) using partial derivative and gradients, which I don't know how to use. I'd like to find a more elementary proof. I know that the fibers of $f$ (or $g$) are confocal ellipses (or hyperbolas) which partition the plane. So the points "inside" an ellipse have smaller $f$ than the points on the ellipse and points "between" the branches of a hyperbola have greater $g$ than those ones on the hyperbola.
Even if it made sense, I can't prove that a tangent line to a conic stays "outside"/"between" that conic (except for the tangent point)... A line tangent to a conic intersect it only at one point (because it is algebraic curve of second degree), but I can't say more so far.
 A: Let's prove first of all that the bisector of the external angle at $T$ of triangle $ABT$ has only point $T$ in common with the ellipse. Consider any point $P\ne T$ on the bisector and point $B'$, reflection of $B$ around the bisector (see diagram below). Notice that $ATB'$ are aligned and $APB'$ form a triangle, so by triangle inequality you have:
$$
AP+BP=AP+PB'> AB'=AT+TB,
$$
proving that $P$ is outside the ellipse.

Now we want to prove that any other line $r$ through $T$, different from the bisector considered above, has at least one point inside the ellipse. 
That is obvious if foci $A$ and $B$ lie on different sides of $r$, because in that case there is a point of segment $AB$ belonging to $r$.
If $A$ and $B$ are on the same side of $r$ (see diagram below), let $B'$ be the reflection of $B$ around $r$ and $P$ the point where $BB'$ intersects $r$. 
Notice that $ATB'$ are not aligned, for in that case $r$ would be the bisector of the external angle at $T$ of triangle $ABT$. By triangle inequality we then have
$$
AP+BP=AB'<AT+TB'=AT+BT,
$$
which shows that $P$ is inside the ellipse.

A: (1) Assume that $$ F'(-k,0),\ F(k,0) $$
and $$ PF+PF'=2a,\ a>k>0 $$ where $PF$ is a distance from $P$ to $F$
So by direct computation
\begin{eqnarray*}
 2a&=& \sqrt{ (x-k)^2+ y^2} + \sqrt{(x+k)^2+y^2} \\
  \sqrt{ (x-k)^2+ y^2}  &=& 2a-  \sqrt{(x+k)^2+y^2}
  \\  (x-k)^2+ y^2 &=& 4a^2 + (x+k)^2+y^2 -4a   \sqrt{ (x+k)^2+
y^2}  \\
kx  +a^2 &=& a \sqrt{ (x+k)^2+ y^2} \\
(a^2-k^2)x^2 + a^2 y^2 &=& a^2(a^2-k^2)\end{eqnarray*}
Hence we have
$$
\frac{x^2}{a^2} + \frac{y^2}{a^2-k^2} =1,\ PF+PF'=2a,\ F,\ F'=
(\pm k,0)$$
(2) Tangent Line of Ellipse : If $Q$ is another point in
ellipse, and if line segment $tP+(1-t)Q$ for $t\in [0,1]$ is in a
region enclosed by ellipse, then ellipse is in one side of a
tangent line
Note that \begin{align*} &F (tP+(1-t)Q) + F'(tP+(1-t)Q) \\&\leq  tFP
+ (1-t) FQ + tF'P+ (1-t)F'Q \\&=2a \end{align*}
A: The interior of an ellipse is bounded. If a line $\ell$ pass through a point $p$ in the interior of an ellipse $\Gamma$, one may see that $\ell$ is divided into three parts: the point $p$, and two rays with endpoint $p$. Each of the ray contains some point in the interior of $\Gamma$, as well as some point in the exterior of $\Gamma$(since the ray is not bounded, hence cannot be contained in a bounded region). So the ray must intersect with $\Gamma$. Since the two rays have no common point except $p$, so we get two different intersection point of $\ell$ and $\Gamma$, a contradiction.  

In the above proof we do not have to invoke Jordan's Curve Theorem to say the interior and exterior of $\Gamma$. Suppose $\Gamma$ is defined by $f(X)=K$. We just define the interior if $\Gamma$ as the set of points $X$ satisfying $f(X)<K$ and exterior as $f(X)>K$. It's easy to prove that the interior is bounded.
