15 The parabola $y=x^{2}$ intersects with the line $l$ passing through the point $P(-1,-1)$ at points $P_{1}$ and $P_{2}$.
(1) Find the range of the slope $k$ of the line $l$;
(2) Find the equation of the trajectory of point $Q$ that satisfies the condition $\frac{1}{P P_{1}}+\frac{1}{P P_{2}}=\frac{2}{P Q}$ on the segment $P_{1} P_{2}$.
(15 (1) The equation of the line $l$ is $y+1=k(x+1)$, and when combined with the parabola equation $y=x^{2}$, we get 
$$
\left\{\begin{array}{l}y=x^{2}, \\ y+1=k(x+1),\end{array}\right.
$$
Eliminating $y$ gives
$$
x^{2}=k(x+1)-1 \text {, }
$$

which simplifies to
$$
x^{2}-k x-(k-1)=0,
$$

From
$$
\Delta=(-k)^{2}+4(k-1)>0,
$$

we solve to get $k>-2+2 \sqrt{2}$ or $k<-2-2 \sqrt{2}$.

(2) Let the coordinates of points $P_{1}$ and $P_{2}$ be $\left(x_{1}, x_{1}^{2}\right)$ and $\left(x_{2}, x_{2}^{2}\right)$, respectively, and the coordinates of point $Q$ be $(x, y)$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=k, \\
x_{1} x_{2}=-(k-1),
\end{array}\right.
$$

Since point $Q$ lies on the segment $P_{1} P_{2}$, $x_{1}+1, x_{2}+1, x+1$ have the same sign. Therefore,
$$
\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}=\frac{2}{x+1} \text {. }
$$

Thus,
$$
\begin{array}{c}
x=2 \frac{x_{1} x_{2}+x_{1}+x_{2}+1}{x_{1}+x_{2}+2}-1=\frac{2-k}{k+2}, \\
y=k(x+1)-1=k \cdot\left(\frac{2-k}{k+2}+1\right)-1=\frac{3 k-2}{k+2},
\end{array}
$$

From (1), we have $k=\frac{2-2 x}{x+1}$. Substituting into (2) gives
$$
y=\frac{3 \frac{2-2 x}{x+1}-2}{\frac{2-2 x}{x+1}+2}=1-2 x,
$$

which simplifies to $2 x-y+1=0$.
Since $k>-2+2 \sqrt{2}$ or $k<-2-2 \sqrt{2}$, the range of $x=\frac{4}{k+2}-1$ is $-\sqrt{2}-1<x<\sqrt{2}-1$ and $x \neq-1$. Therefore, the equation of the trajectory of point $Q$ is $2 x-y+1=0(-\sqrt{2}-1<x<\sqrt{2}-1$ and $x \neq-1)$.