= Solution
{c}
A <hyperbola> with <focus of a conic> $F$, <directrix of a conic> $D$ not containing $F$, and <eccentricity> $e>1$ is the <locus>
$$
\{P:\operatorname{dist}(P,F)=e\operatorname{dist}(P,D)\}.
$$
Put $c=\sqrt{a^2+b^2}$ and $e=c/a>1$. For $F=(c,0)$ and $D=\{x=a^2/c\}$, the defining equality becomes
$$
\sqrt{(x-c)^2+y^2}=\frac ca\left|x-\frac{a^2}{c}\right|.
$$
Both sides are nonnegative, so squaring is reversible. Expanding and cancelling the linear terms gives
$$
x^2-2cx+c^2+y^2=\frac{c^2}{a^2}x^2-2cx+a^2
\quad\Longleftrightarrow\quad
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.
$$
Thus the two matched <focus of a conic> and <directrix of a conic> choices, related by reflection, are
$$
\boxed{F_\pm=(\pm\sqrt{a^2+b^2},0),\qquad
D_\pm:\ x=\pm\frac{a^2}{\sqrt{a^2+b^2}},\qquad
e=\sqrt{1+\frac{b^2}{a^2}}.}
$$
The signs in each pair must agree.
On the <hyperbola>, $y=\pm(b/a)\sqrt{x^2-a^2}$. As $|x|\to\infty$,
$$
|x|-\sqrt{x^2-a^2}=\frac{a^2}{|x|+\sqrt{x^2-a^2}}\longrightarrow0.
$$
Hence the perpendicular <distance> to one of the two <straight lines> tends to zero on each end of a branch. The <asymptotes> are therefore
$$
\boxed{y=\frac ba x\quad\text{and}\quad y=-\frac ba x.}
$$
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