15. The line $y=k x+1$ intersects the left branch of the hyperbola $x^{2}-y^{2}=1$ at points $A$ and $B$. The line $l$ passes through the point $(-2,0)$ and the midpoint of $A B$. Find the range of the y-intercept $b$ of the line $l$.
15. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. By combining the equations of the line and the hyperbola, we get 
$$
\left\{\begin{array}{l}y=k x+1, \\ x^{2}-y^{2}=1 .\end{array}\right.
$$
Simplifying, we obtain
$$
\left(k^{2}-1\right) x^{2}+2 k x+2=0 .
$$

From the problem, we know that equation (1) has two negative roots, therefore,
$$
\left\{\begin{array}{l}
\Delta=4 k^{2}-8\left(k^{2}-1\right)>0 \\
x_{1}+x_{2}=-\frac{2 k}{k^{2}-1}0 .
\end{array}\right.
$$

Solving, we get $1<k<\sqrt{2}$.
Then $y_{1}+y_{2}=k\left(x_{1}+x_{2}\right)+2$
$$
=-\frac{2 k^{2}}{k^{2}-1}+2=-\frac{2}{k^{2}-1} \text {. }
$$

Therefore, the midpoint of $A B$ is $\left(-\frac{k}{k^{2}-1},-\frac{1}{k^{2}-1}\right)$.
Thus, the equation of line $l$ is
$$
y=-\frac{1}{2 k^{2}-k-2}(x+2),
$$

with the y-intercept $b=-\frac{2}{2 k^{2}-k-2}$.
When $1<k<\sqrt{2}$,
$$
2 k^{2}-k-2=2\left(k-\frac{1}{4}\right)^{2}-\frac{17}{8} \text {, }
$$

its range is $(-1,2-\sqrt{2})$.
Hence, the range of $b$ is
$$
(-\infty,-2-\sqrt{2}) \cup(2,+\infty) \text {. }
$$