Some clarification needed 🙏🏻

[humble tusk]

<@&286206848099549185> help pls

[harsh pebble]

The distance between F1 and F2 is 2c, right?. So the shortest distance you can travel from F1 to F2 is straight horizontaly on the x-axis.
Lets asume P is not on this horizontal line. If that is so, we can construct the triangle F1F2P, where 2 of the sides PF1 and PF2 are the distance 2k, and the side F1F2 is 2c. We can agree, that there is no triangle, where the length of 2 sides added is shorter than the third side, thus 2k must be greater than 2c, and with that k>c.
Now lets asume P is on the horizontal line. If that is the case, the y-value of P is automaticaly 0, and the x-value is less than or equal to c. Since y is 0, the equasion x²/a² + y²/b² = 1 turns to x²/a² = 1. So depending on the c and a value, there might be 2 points on the horizontal distance, where P can be placed, with which 2k = 2c is True. But 2c can never be greater than 2k, as 2c is the shortest distance between F1 and F2.

Now typing this out I'm starting to think this is not the answer you are looking for, so sorry about that.

[tacit wharf]

The alternatives to $k$ being positive are:
\begin{itemize} \item $k=c$ \item $k<c$ \end{itemize}
If $k=c$ then there would only all points $P$ in the set ${(x,0):-c\leq x\leq c}$ fulfill $PF_1+PF_2=2k$. This line segment could hardly describe an ellipse.
On the other hand, if $k<c$ then there are no points $P$ that fulfill $PF_1+PF_2=2k$ since the minimum distance between two points is given by the line through these two points, this gives a minimum distance of $2c$ between $F_1$ and $F_2$. This means that there is no way that a point $P$ could fulfill $PF_1+PF_2=2k$ if $k<c$.

To sum up, if $k=c$ then you have a set of points which would arguably not describe an ellipse. If $k<c$ the no points $P$ exist that fulfill $PF_1+PF_2=2k$ since the minimum distance between $F_1$ and $F_2$ is $2c$.

[vocal rain]

Crystopher

[humble tusk]

ohhh okay