Minimum distance from a point to a parabola

Text

Find the minimum distance between the point $P(3,0)$ and the parabola $y=x^{2}$.

Solution

A point of the parabola is $(x,x^{2})$; squared distance $D(x)=(x-3)^{2}+x^{4}=x^{4}+x^{2}-6x+9$. $D'(x)=4x^{3}+2x-6$; for $x=1$: $4+2-6=0$. Since $D''(x)=12x^{2}+2>0$ it is a minimum. Nearest point $(1,1)$; $D(1)=5$, so $d=\sqrt5\approx2.236$. $\boxed{x=1,\ (1,1);\quad d_{\min}=\sqrt5\approx2.236}$