6. Given the equation of $\odot O$ is $x^{2}+y^{2}=1$, and point $A(1,0)$. Take a moving point $B$ on $\odot O$, and let point $P$ satisfy $\overrightarrow{A P}=\lambda \overrightarrow{O B}(\lambda \in \mathbf{R})$ and $\overrightarrow{A P} \cdot \overrightarrow{A B}=1$. Then the equation of the trajectory of point $P$ is $\qquad$ .
6. $y^{2}=2 x-1$.

Let point $P(x, y)$. Then by $\overrightarrow{A P}=\lambda \overrightarrow{O B}(\lambda \in \mathbf{R})$, we get $B(k(x-1), k y)\left(\lambda=\frac{1}{k}\right)$.
Substituting the coordinate formula into $\overrightarrow{A P} \cdot \overrightarrow{A B}=1$, we get
$$
k=\frac{x}{(x-1)^{2}+y^{2}} \text {. }
$$

Since point $B$ is on $\odot O$, then
$$
k^{2}(x-1)^{2}+k^{2} y^{2}=1 \text {. }
$$

Combining equations (1) and (2) and eliminating $k$, we get $y^{2}=2 x-1$.