6. The endpoints of a line segment $AB$ of fixed length 3 move on the parabola $y^{2}=x$, and $M$ is the midpoint of segment $AB$. Find the shortest distance from $M$ to the $y$-axis.
(1987 National College Entrance Examination)
6. Let $M\left(x_{0}, y_{0}\right), A\left(x_{0}+d_{1}, y_{0}+d_{2}\right), B\left(x_{0}-d_{1}, y_{0}-d_{2}\right)$, then $\left(y_{0}+d_{2}\right)^{2}=x_{0}+d_{1},\left(y_{0}-d_{2}\right)^{2}=$ $x_{0}-d_{1}$. Subtracting these two equations yields $y_{0}=\frac{d_{1}}{2 d_{2}}$, and adding them gives $x_{0}=y_{0}^{2}+d_{2}^{2}=\frac{d_{1}^{2}}{4 d_{2}^{2}}+d_{2}^{2}$. Also, $|P Q|^{2}=\left(2 d_{1}\right)^{2}+\left(2 d_{2}\right)^{2}=$ $3^{2}$, so $d_{2}^{2}=\frac{1}{4}\left(9-4 d_{1}^{2}\right)$. Therefore, $x_{0}=\frac{d_{1}^{2}}{9-4 d_{1}^{2}}+\frac{1}{4}\left(9-4 d_{1}^{2}\right)=-\frac{1}{4}+\frac{9}{4\left(9-4 d_{1}^{2}\right)}+\frac{9-4 d_{1}^{2}}{4} \geqslant \frac{5}{4}$, with equality if and only if $d_{1}= \pm \frac{\sqrt{6}}{2}$, in which case $\left(x_{0}\right)_{\text {min }}=\frac{5}{4}$, meaning the shortest distance from $M$ to the $y$-axis is $\frac{5}{4}$.