jmc

Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix.

Since the parabola $y=x^2$ is symmetric about the $y$-axis, the focus is at a point of the form $(0,f).$ Let $y=d$ be the equation of the directrix.



Let $(x,x^2)$ be a point on the parabola $y=x^2.$ Then $$P{F}^2=x^2+(x^2-f{)}^2$$and $P{Q}^2=(x^2-d{)}^2.$ Thus, $$x^2+(x^2-f{)}^2=(x^2-d{)}^2.$$Expanding, we get $$x^2+x^4-2f{x}^2+f^2=x^4-2d{x}^2+d^2.$$Matching coefficients, we get $$\begin{array}{rl}1-2f& =-2d,\\ f^2& =d^2.\end{array}$$From the first equation, $f-d=\frac{1}{2}.$ Since $f^2=d^2,$ $f=d$ or $f=-d.$ We cannot have $f=d,$ so $f=-d.$ Then $2f=\frac{1}{2},$ so $f=\frac{1}{4}.$

Thus, the focus is $\boxed{\left(0,\frac{1}{4}\right)}.$