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 tangency point is that point $P$ on the given line having the minimum sum of distances from foci $A=(1,-1)$ and $B=(2,-1)$. But it is well known how to find such a point: reflect point $B$ about the line, to get $B'=(6,3)$, and $P$ is then the intersection between the given line and line $AB'$. A simple computation gives then $P=(34/9,11/9)$.

