Given the lines $1_{1}: a x-k y+k=0$;  
$1_{2}: \mathrm{kx}-\mathrm{y}-1=0$, where $\mathrm{a} \neq 0$, $\mathrm{a}$ is a constant, and $\mathrm{k}$ is a parameter. Coordinates of the point and eccentricity:  
(2) When $a>0, y \geqslant 1$, find the minimum distance from point $P(x, y)$ on the trajectory to point $A(0, b)$.
Solve (1) From $\left\{\begin{array}{l}a x-k y+k=0, \\ k x-y-1=0\end{array}\right.$, eliminating the parameter $k$ (using substitution method) yields $-a x^{2}+y^{2}=1$. When $a>0$, the curve is a hyperbola. Converting it to the standard form gives: $-\frac{x^{2}}{\frac{1}{a}}+y^{2}=1, c=\sqrt{1+\frac{1}{a}}$.

Thus, the coordinates of the foci are $\left(0, \pm \sqrt{1+\frac{1}{a}}\right)$, and the eccentricity $e=\sqrt{1+\frac{1}{a}}$.
When $a=-1$, it is a circle.
When $a<0$,
$\therefore \mathrm{c}=\sqrt{-\frac{1}{a}-1}$, the foci are $\left( \pm \sqrt{-\frac{1}{a}-1}, 0\right)$, and the eccentricity $e=\frac{\sqrt{-\frac{1}{a}} 1}{\sqrt{-\frac{1}{a}}}=\sqrt{1+a}$.
When $a>0$, the trajectory is a hyperbola, and since $\mathrm{y} \geqslant 1$, we take the upper half-plane branch.
Let $|P A|=1$, then
$1^{2}=x^{2}+(y-b)^{2}$
$$
=\frac{y^{2}-1}{a}+
$$
$\left(y^{2}-2 b y+b^{2}\right)$
$$
=\frac{a+1}{a}\left(y-\frac{a b}{a+1}\right)^{2}+\frac{a b^{2}-a-1}{a(a+1)}
$$
$$
=f(y) . \quad(y \geqslant 1, a>0)
$$

Here $f(y)$ is a quadratic function of $y$, with the axis of symmetry given by $y=\frac{a b}{a+1}$. When the axis of symmetry passes through the domain, i.e., $\frac{a b}{a+1} \geqslant 1 \quad\left(b \geqslant \frac{a+1}{a}\right)$, then
$$
l^{2} \text { minimum }=\mathrm{f}\left(\frac{\mathrm{ab}}{\mathrm{a}+1}\right)=\frac{a b^{2}-a-1}{a(a+1)} \text {. }
$$

Thus, $1 \text { minimum }=\sqrt{\frac{a b^{2}-a-1}{a(a+1)}}$. $\left(\right.$ Clearly, $a b^{2}-a$ $-1>0$ )

When the axis of symmetry does not pass through the domain, i.e., $\frac{a b}{a+1}<1$ $<\frac{a+1}{a}$, then $1^{2}$ minimum $=\mathrm{f}(1)=(1-\mathrm{b})^{2}$. Thus, $1_{\text {minimum }}=|1-b|$, so
$1_{\text {minimum }}=\left\{\begin{array}{ll}\sqrt{\frac{a b^{2}-a-1}{a(a+1)}}, & b \geqslant \frac{a+1}{a} \text { when. } \\ |1-b|, & b<\frac{a+1}{a} \text { when. }\end{array}\right.$