One focus $A$ and two points of an ellipse $C$ and $D$ are given. Let $f$ be a tanget line at $C$. Construct the other focus.

Question

Obviously, the other focus $B$ is somewhere on the line $CE'$ where $E'=s_f(E)$. Now I need one more condition.

Answer 1

Construct $A'$, reflection of $A$ about tangent $f$. Focus $B$ lies on ray $A'C$. On that ray construct then $D'$ such that $A'D'=AD$. The second focus $B$ of the ellipse lies on the perpendicular bisector of $DD'$ and can thus be constructed.

PROOF.

Let $B$ be the second focus of the ellipse, on ray $A'C$. We have: $$ AD+DB=AC+CB=A'B=A'D'+D'B. $$ But $AD=A'D'$, hence $DB=D'B$, QED.

EDIT.

In the proof I assumed point $B$ to lie on ray $A'C$ and after $C$. If, on the contrary, the intersection $B$ between line $A'C$ and the perpendicular bisector of $DD'$ lies on ray $CA'$, then no ellipse as described can exist, but the hyperbola with foci $A$, $B$ and passing through $C$ is also tangent to $f$ and passes through $D$.