An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse. 
An ellipse has foci at $(1, -1)$ and $(2, -1)$ and tangent $x+y-5=0$. Find the point where the tangent touches the ellipse.

Here is a procedure how to do it analiticaly.

*

*If $T(x_0,y_0)$ is a touching point, then $x_0+y_0=5$

*The equation of ellipse is $${(x_0-{3\over 2})^2\over a^2} +{(y_0+1)^2 \over b^2}=1$$

*Since $2e=1$ we have $a^2-b^2 = {1\over 4}$

*Since the slope of tangent is $-1$ we have $${2(x_0-{3\over 2})\over a^2} -{2(y_0+1)\over b^2}=0$$
And now we have to solve this tedious system. How to do it more geometrical?

The ellipse divides the plane into three regions.

*

*the exterior, where the sum of the distances to the foci is greater than $c$.

*the ellipse itself, where the sum of the distances to the foci is exactly $c$.

*the interior, where the sum of the distances to the foci is smaller than $c$.

Parametrize the tangent as $x=t,y=5-t$. Thus the sum of the distances to the foci along the tangent has unique global minimum at the tangency point.
Explicitly, the function
\begin{align}
f(t):={}&\sqrt{(t-1)^2+(5-t+1)^2}+
\sqrt{(t-2)^2+(5-t+1)^2}
\\
={}&
\sqrt{(t-1)^2+(6-t)^2}+
\sqrt{(t-2)^2+(6-t)^2}
\end{align}
has global minimum at $t=\frac {34}9$, yielding the tangency point $(\frac {34}9,\frac{11}9)$. This can be found by differentiating $f(t)$ and imposing $f'(t)=0$, which gives
$$
\frac{4 \sqrt{2} (t-4)}{\sqrt{(t-8) t+20}}=-\frac{8 t-28}{\sqrt{2 (t-7) t+37}}
$$
which can be squared to a quadratic equation
$$
9 t^2-88 t+204=0\Rightarrow (t-6 ) ( 9 t-34)=0,
$$
with roots $6,\frac {34}9$, the latter being the actual solution of the original equation before squaring.
