6. Given a fixed point $P(0,1)$, a moving point $Q$ satisfies that the perpendicular bisector of line segment $P Q$ is tangent to the parabola $y=x^{2}$. Then the equation of the locus of point $Q$ is $\qquad$ .
6. $(2 y-1) x^{2}+2(y+1)(y-1)^{2}=0\left(-1 \leqslant y<\frac{1}{2}\right)$.

Let the perpendicular bisector $l$ of $P Q$ be tangent to the parabola $y=x^{2}$ at the point $\left(t, t^{2}\right)$, with the tangent vector $(1,2 t)$.
Then the line $l: y=2 t(x-t)+t^{2}$.
Let $Q(x, y)$.
From $P Q \perp l$ and the midpoint of $P Q$ lying on the line $l$, we have
$$
\left\{\begin{array}{l}
x+2 t(y-1)=0, \\
\frac{1}{2}(y+1)=t x-t^{2} .
\end{array}\right.
$$

From equation (1), we solve for $t=\frac{x}{2-2 y}$.
Substituting into equation (2) yields the trajectory equation of point $Q$ as
$$
(2 y-1) x^{2}+2(y+1)(y-1)^{2}=0\left(-1 \leqslant y<\frac{1}{2}\right) \text {. }
$$