The closest point on a parabola

The function. For the point $(x,x^2)$,
$$d^2=x^2+(x^2-1)^2=x^4-x^2+1.$$

Substitution. Put $u=x^2\ge0$. Then $d^2=u^2-u+1$, a quadratic in $u$ minimised at $u=\frac12$ (the derivative $2u-1$ vanishes there, and is negative before and positive after). Hence $x=\pm\frac1{\sqrt2}$ and
$$d^2=\frac14-\frac12+1=\frac34,\qquad d=\frac{\sqrt3}2\approx0.866.$$

Answer. The closest points are $\left(\pm\frac1{\sqrt2},\frac12\right)$, at distance $\frac{\sqrt3}2$.

Remark. The vertex $(0,0)$ is at distance $1$, which is *not* the closest: it is a local maximum of $d^2$ on the parabola, since $d^2=x^4-x^2+1$ decreases when $|x|$ increases slightly from $0$. At the optimum the segment from $A$ is perpendicular to the tangent of the parabola.