Answer
As $P$ approaches the origin $Q$ approaches the point $(0, \frac{1}{2})$
Work Step by Step
Let the x-coordinate of $P$, $x_P = a$, since P lies on the parabola $y = x^{2}$ $P$ must have coordinates $(a,a^{2})$.
The line $OP$ must be in the form $y=mx$, since it goes through the origin.
$m = \frac{Δy}{Δx} = \frac{a^{2}}{a} = a$
So $OP: y = ax$
Let $l$ be the perpendicular bisector of $OP$. $l$ must have the form $l:y = cx + d$. Since it is perpendicular to $OP$, $c = -\frac{1}{a}$
The point halfway between $O$ and $P$ will be $(\frac{1}{2}a,\frac{1}{2}a^{2})$
Since $l$ passes through this point we can say that
$y = -\frac{1}{a}x+d$
$\frac{1}{2}a^{2} = -\frac{1}{a}(\frac{1}{2}a) + d$
$\frac{1}{2}a^{2} = -\frac{1}{2} + d$
$d = \frac{1}{2}a^{2}+\frac{1}{2}$
Which gives $l:y=-\frac{1}{a}x+\frac{1}{2}a^{2}+\frac{1}{2}$
The x-coordinate of $Q$ must be $0$ and $Q$ must lie on $l$ thus we get
$y_Q = -\frac{1}{a}(0)+\frac{1}{2}a^{2}+\frac{1}{2} = \frac{1}{2}a^{2}+\frac{1}{2}$
Thus $Q$ must have coordinates $(0,\frac{1}{2}a^{2}+\frac{1}{2})$
As $P$ approaches the origin $a \to 0$
Thus we compute
$\lim\limits_{a \to 0}(\frac{1}{2}a^{2}+\frac{1}{2})$
$= \frac{1}{2}$
So we get coordinates for $Q$ $(0,\frac{1}{2})$ as $P$ approaches the origin.
