6. Let the points be
$$
A\left(a, a+\frac{1}{2}\right), B\left(a+1, a+\frac{3}{2}\right) \text {, }
$$

A moving point $P$ is such that its distance to point $M(1,0)$ is 1 unit greater than its distance to the $y$-axis, and its trajectory forms a curve $C$. The line segment $AB$ intersects curve $C$. Then the range of values for $a$ is $\qquad$
6. $\left[\frac{1}{2}-\sqrt{2}, \frac{3}{2}-\sqrt{2}\right] \cup\left[\frac{1}{2}+\sqrt{2}, \frac{3}{2}+\sqrt{2}\right]$. From the given conditions, the equation of curve $C$ is $y^{2}=4 x$.
To ensure that segment $A B$ intersects with curve $C$, it is sufficient for points $A$ and $B$ to be on opposite sides of the parabola. Therefore, let
$$
\begin{array}{l}
F(x, y)=y^{2}-4 x . \\
\text { Then } F\left(a, a+\frac{1}{2}\right) F\left(a+1, a+\frac{3}{2}\right) \\
=\left(a^{2}-3 a+\frac{1}{4}\right)\left(a^{2}-a-\frac{7}{4}\right) \leqslant 0 .
\end{array}
$$

We can find that
$$
a \in\left[\frac{1}{2}-\sqrt{2}, \frac{3}{2}-\sqrt{2}\right] \cup\left[\frac{1}{2}+\sqrt{2}, \frac{3}{2}+\sqrt{2}\right]
$$