9. Given $O$ as the origin, $N(1,0)$, and $M$ as a moving point on the line $x=-1$, the angle bisector of $\angle M O N$ intersects the line $M N$ at point $P$, and the locus of point $P$ is curve $E$.
(1) Find the equation of curve $E$;
(2) A line $l$ with slope $k$ passes through the point $Q\left(-\frac{1}{2},-\frac{1}{2}\right)$. If line $l$ intersects curve $E$ at exactly one point, find the range of values for $k$.


(1) Let $P(x, y), M(-1, t)$, it is easy to know $0 \leq x<1$. Since $O P$ bisects $\angle M O N$, we have
$$
\overrightarrow{M P}=\frac{|O M|}{|O N|} \overrightarrow{P N}=\sqrt{1+t^{2}} \overrightarrow{P N}
$$

Therefore,
$$
\begin{array}{l}
x+1=\sqrt{1+t^{2}}(1-x) \\
y-t=\sqrt{1+t^{2}}(0-y)
\end{array}
$$

From (1) and (2), we get $t=\frac{2 y}{1-x}$, substituting into (1) yields $\frac{x+1}{1-x}=\sqrt{1+\left(\frac{2 y}{1-x}\right)^{2}}$, simplifying we get the equation of curve $E$ as $y^{2}=x(0 \leq x<1)$;
(2) Let $A(1,1), B(1,-1)$, then $k_{Q A}=1, k_{Q B}=-\frac{1}{3}$.
The equation of line $l$ is $y+\frac{1}{2}=k\left(x+\frac{1}{2}\right)$, combining with the parabola equation $y^{2}=x$, eliminating $x$ we get $k y^{2}-y+\frac{1}{2}(k-1)=0$. When line $l$ is tangent to the parabola $y^{2}=x$ at point $T$, $\Delta=1-2 k(k-1)=0$, solving we get $k_{1,2}=\frac{1 \pm \sqrt{3}}{2}$: when $k=k_{1}=\frac{1+\sqrt{3}}{2}$, $y_{r}=\frac{\sqrt{3}-1}{2}$, the tangent point $T$ is on curve $E$; when $k=k_{2}=\frac{1-\sqrt{3}}{2}$, $y_{r}=-\frac{\sqrt{3}+1}{2}$, the tangent point $T$ is not on curve $E$.

If line $l$ intersects curve $E$ at exactly one point, then $k_{Q B}<k \leq k_{Q A}$ or $k=\frac{1+\sqrt{3}}{2}$, hence the range of $k$ is
$$
\left(-\frac{1}{3}, 1\right] \cup\left\{\frac{1+\sqrt{3}}{2}\right\} .
$$