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. Solution: By combining the line equation $y=k x+1$ with the hyperbola equation $x^{2}-y^{2}=1$, we get $\left\{\begin{array}{l}y=k x+1 \\ x^{2}-y^{2}=1\end{array}\right.$, which simplifies to $\left(k^{2}-1\right) x^{2}+2 k x+2=0$ (1)

From the problem, we know that equation (1) has two negative roots, so $\left\{\begin{array}{l}\Delta=4 k^{2}-8\left(k^{2}-1\right)>0 \\ x_{1}+x_{2}=-\frac{2 k}{k^{2}-1}<0\end{array}\right.$, solving this gives $1<k<\sqrt{2}$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then we have $x_{1}+x_{2}=-\frac{2 k}{k^{2}-1}$,
$$
y_{1}+y_{2}=k\left(x_{1}+x_{2}\right)+2=-\frac{2 k^{2}}{k^{2}-1}+2=-\frac{2}{k^{2}-1}
$$

Thus, the midpoint of $A B$ is $\left(-\frac{k}{k^{2}-1},-\frac{1}{k^{2}-1}\right)$,
Therefore, the equation of line $l$ is $y=\frac{-1}{2 k^{2}-k-2}(x+2)$, and its y-intercept $b=\frac{-2}{2 k^{2}-k-2}$. When $1<k<\sqrt{2}$, $2 k^{2}-k-2=2\left(k-\frac{1}{4}\right)^{2}-\frac{17}{8}$, its range is $(-1,2-\sqrt{2})$, so the range of $b=\frac{-2}{2 k^{2}-k-2}$ is $(-\infty,-2-\sqrt{2}) \cup(2,+\infty)$.