14. (20 points) Given a constant $a>0$, vectors $\boldsymbol{p}=(1, 0)$, $\boldsymbol{q}=(0, a)$, a line passing through the fixed point $M(0, -a)$ with direction vector $\lambda \boldsymbol{p} + \boldsymbol{q}$ intersects with a line passing through the fixed point $N(0, a)$ with direction vector $\boldsymbol{p} + 2\lambda \boldsymbol{q}$ at point $R$, where $\lambda \in \mathbb{R}$.
(1) Find the equation of the trajectory of point $R$;
(2) Let $a=\frac{\sqrt{2}}{2}$, and a line $l$ passing through $F(0, 1)$ intersects the trajectory of point $R$ at points $A$ and $B$. Find the range of $\boldsymbol{F A} \cdot \boldsymbol{F B}$.
14. (1) Let $R(x, y)$, then
$\boldsymbol{M R}=(x, y+a), \boldsymbol{N R}=(x, y-a)$.
Also, $\lambda \boldsymbol{p}+\boldsymbol{q}=(\lambda, a), \boldsymbol{p}+2 \lambda \boldsymbol{q}=(1,2 \lambda a)$, and
$\boldsymbol{M R} / /(\lambda p+q), N R / /(p+2 \lambda q)$,
thus $\left\{\begin{array}{l}\lambda(y+a)=a x, \\ y-a=2 \lambda a x .\end{array}\right.$
Eliminating the parameter $\lambda$, we get the trajectory equation of point $R$ as
$(y+a)(y-a)=2 a^{2} x^{2}$,
which simplifies to $y^{2}-2 a^{2} x^{2}=a^{2}$ (excluding the point $(0,-a)$ ).
(2) Since $a=\frac{\sqrt{2}}{2}$, the trajectory equation of point $R$ is
$2 y^{2}-2 x^{2}=1$ (excluding the point $\left(0,-\frac{\sqrt{2}}{2}\right)$ ).
If the slope of $l$ does not exist, its equation is $x=0$, and $l$ intersects the hyperbola at one point $A\left(0, \frac{\sqrt{2}}{2}\right)$.

If the slope of $l$ exists, let its equation be $y=k x+1$. Substituting into $2 y^{2}-2 x^{2}=1$, we get
$2\left(k^{2}-1\right) x^{2}+4 k x+1=0$.
From $\left\{\begin{array}{l}k^{2}-1 \neq 0, \\ \Delta=16 k^{2}-8\left(k^{2}-1\right)>0,\end{array}\right.$ we solve to get $k \neq \pm 1$.
Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$, then $x_{1} x_{2}=\frac{1}{2\left(k^{2}-1\right)}$.
Thus, $F A \cdot F B=\left(x_{1}, y_{1}-1\right) \cdot\left(x_{2}, y_{2}-1\right)$
$=\left(x_{1}, k x_{1}\right) \cdot\left(x_{2}, k x_{2}\right)=x_{1} x_{2}+k^{2} x_{1} x_{2}$
$=\frac{k^{2}+1}{2\left(k^{2}-1\right)}=\frac{1}{2}\left(1+\frac{2}{k^{2}-1}\right)$.
When $-11$ or $k0, F A \cdot F B>\frac{1}{2}$.
In summary, we have
$$
\boldsymbol{F A} \cdot \boldsymbol{F B} \in\left(-\infty,-\frac{1}{2}\right] \cup\left(\frac{1}{2},+\infty\right) .
$$