jmc

Let $A=(a,a^2)$ and $C=(c,c^2).$ Since $\overline{OA}$ and $\overline{OC}$ are perpendicular, the product of their slopes is $-1$: $$\frac{a^2}{a}\cdot \frac{c^2}{c}=-1.$$Thus, $ac=-1.$



As a rectangle, the midpoints of the diagonals coincide. The midpoint of $\overline{AC}$ is $$\left(\frac{a+c}{2},\frac{a^2+c^2}{2}\right),$$so $B=(a+c,a^2+c^2).$

Let $x=a+c$ and $y=a^2+c^2.$ We want a relationship between $x$ and $y$ in the form of $y=p{x}^2+qx+r.$ We have that $$x^2=(a+c{)}^2=a^2+2ac+c^2=a^2+c^2-2=y-2,$$so the fixed parabola is $\boxed{y=x^2+2}.$