16. Given a constant $a>0$, vector $m=(0, a)$, $n=(1,0)$, a line passing through the fixed point $A(0,-a)$ with the direction vector $m+\lambda n$ intersects with a line passing through the fixed point $B(0, a)$ with the direction vector $n+2 \lambda m$ at point $P$, where $\lambda \in \mathbf{R}$. Then the equation of the trajectory of point $P$ is $\qquad$
16. $y^{2}-a^{2}=2 a^{2} x^{2}$.

Let point $P(x, y)$, then $A P=(x, y+a), B P=(x, y-a)$.
Also, $\boldsymbol{n}=(1,0), \boldsymbol{m}=(0, a)$, so
$$
\boldsymbol{m}+\lambda \boldsymbol{n}=(\lambda, a), \boldsymbol{n}+2 \lambda \boldsymbol{m}=(1,2 \lambda a) \text {. }
$$

By the given condition, vector $A P$ is parallel to vector $\boldsymbol{m}+\lambda \boldsymbol{n}$, hence
$$
\lambda(y+a)=a x \text {. }
$$

Also, vector $B P$ is parallel to vector $\boldsymbol{n}+2 \lambda \boldsymbol{m}$, hence
$$
y-a=2 \lambda a x \text {. }
$$

Solving the two equations simultaneously and eliminating the parameter $\lambda$, we get the trajectory equation of point $P(x, y)$ as $(y+a)(y-a)=2 a^{2} x^{2}$, which simplifies to $y^{2}-a^{2}=2 a^{2} x^{2}$, excluding the point $(0,-a)$.