6. A moving point $P(x, y)$ starts from the origin $O$ and moves along the parabola $y=x^{2}$ to point $A\left(\frac{3}{2}, \frac{9}{4}\right)$, then during the movement, when $P O+P A$ is maximized, the x-coordinate of point $P$ is $\qquad$
6. As shown in the figure

With $O$ and $A$ as two
foci, draw an ellipse that
is tangent to the $\overparen{O A}$ arc (parabolic
arc segment) at $F$. Point ( $\overparen{C .4}$ arc
segment is within the
ellipse or on its boundary
. If there are several tangent
points, choose any one of them). By the definition of an ellipse, it can be deduced that: the sum of the distances from any point inside the ellipse to the two foci is less than the sum of the distances from any point on the ellipse to the two foci (this can be easily proven using the triangle inequality), so for any point $P$ on the $\overparen{O A}$ arc, there must be $P O + P A \leqslant P_{0} O + P_{0} A$, which means $P_{0} O + P_{0} A$ is the maximum value. Below, we find the x-coordinate of $P_{0}$.

Draw the common tangent line $l$ of the ellipse and the parabola at point $P_{0}$. Then, by the optical reflection property of the ellipse: the angles between $O P_{0}$ and $P_{0} A$ with $l$ are equal. Suppose the coordinates of $P_{0}$ are $\left(x_{0}, x_{0}^{2}\right)$, then the slope of $l$ is $2 x_{0}$ (refer to the note of the fifth question in the first part), the slope of $O P_{0}$ is $\frac{x_{0}^{2}}{x_{0}}=x_{0}$, and the slope of $P_{0} A$ is $\frac{\frac{9}{4}-x_{0}^{2}}{\frac{3}{2}-x_{0}}=\frac{3}{2}+x_{0}$.
Using the formula for the angle between two lines, we get
$$
\frac{\left(\frac{3}{2}+x_{0}\right)-2 x_{0}}{1+\left(\frac{3}{2}+x_{0}\right) \cdot 2 x_{0}}=\frac{2 x_{0}-x_{0}}{1+2 x_{0} \cdot x_{0}},
$$

Simplifying, we get $8 x_{0}^{3}+4 x_{0}-3=0$.
Factoring, we get $\left(2 x_{0}-1\right)\left(4 x_{0}^{2}+2 x_{0}+3\right)=0$.
Taking the positive root, we get $x_{0}=\frac{1}{2}$.
In summary, when $P O + P A$ is maximized, the x-coordinate of $P$ is $x=\frac{1}{2}$.