Are the directrix line and focus unique for a conic?

Question

How to prove of disprove the uniqueness of pair of (or single for parabola) directrix line and pair of focal points for a conics?

(Ignore the degenerate cases and the circle.)

Answer 1

Choose a generic parabola $y=ax^2$ (with $a>0$), defined as the locus of points having the same distance from a given focus $F=(0,1/(4a))$ and a given directrix $d$ of equation $y=-1/(4a)$ (the argument for an ellipse or hyperbola is similar but is left to the reader). We must prove that no other pair point/line can be a focus/directrix for the parabola.

Let $F'$, $d'$ be a possible focus/directrix pair. We show at first that $d'$ is parallel to $d$. For if it were not so, we could consider two points $P$, $Q$ on the parabola such that $PQ\parallel d'$, so that $P$ and $Q$ have the same distance from $d'$. That would imply $PF'=QF'$, hence $F'$ would lie on the perpendicular bisector of $PQ$. But a different line $P'Q'\parallel d'$ has a different perpendicular bisector, as you can easily show, parallel to the previous one, which is impossible because $F'$ cannot lie on two different parallel lines. (The case when $d'$ is parallel to the $y$-axis must be treated separately, but it's easily dealt with, as in that case $d'$ would intersect the parabola).

Hence $d'$ must be parallel to $d$. Suppose then $y=-k$ is the equation of $d'$, which entails $F'=(0,k)$. We must have $k>0$, for otherwise $d'$ would intersect the parabola and that leads to a contradiction. Consider then point $S=(\sqrt{k/a},k)$ on the parabola. Its distance from $F'$ is $\sqrt{k/a}$, while its distance from $d'$ is $2k$. Hence it must be $\sqrt{k/a}=2k$, that is $k/a=4k^2$ and $k=1/(4a)$: that proves that $F'=F$ and $d'=d$.