= Solution
{c}
For a fixed <point> $F$ not on a fixed <line> $D$, the <ellipse> with <focus of a conic> $F$, <directrix of a conic> $D$ and <eccentricity> $0<e<1$ is the <locus> of <points> $P$ satisfying
$$
\boxed{\operatorname{dist}(P,F)=e\operatorname{dist}(P,D).}
$$
Set $c=\sqrt{a^2-b^2}$ and $e=c/a$. One matched <focus of a conic> and <directrix of a conic> pair is
$$
\boxed{F=(c,0),\qquad D:\ x=\frac{a^2}{c},\qquad e=\sqrt{1-\frac{b^2}{a^2}}.}
$$
Indeed the squared <distance> condition is
$$
(x-c)^2+y^2=\frac{c^2}{a^2}\left(x-\frac{a^2}{c}\right)^2.
$$
Expanding and cancelling the terms linear in $x$ gives
$$
\left(1-\frac{c^2}{a^2}\right)x^2+y^2=a^2-c^2=b^2,
$$
which is exactly $x^2/a^2+y^2/b^2=1$. Both sides of the unsquared <distance> condition are nonnegative, so squaring introduces no extra <points>. Reflecting across the $y$-axis gives the other matched pair: $F=(-c,0)$ and $D:x=-a^2/c$, with the same <eccentricity>.
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