Shortest distance of the point (0,c) from the parabola $y=x^2$ where $0\le c\le 5$ is

If S is the distance of the point (x,y) on the parabola $y=x^2$ from (0,c) then 

$S=\sqrt{x^2+{(y-c)}^2}=\sqrt{x^2+{(x^2-c)}^2}$

Which is least $\Rightarrow s^2$ is least

$\frac{d{s}^2}{dx}=2x+2(x^2-c)\times 2x=0$

$\Rightarrow x=0$ or $x^2=c-\frac{1}{2}$

Where $x=0,s=c$ when $x^2=c-\frac{1}{2}\ge 0\Rightarrow c\ge \frac{1}{2}$

And $s=\sqrt{c-\frac{1}{2}+\frac{1}{4}}=\sqrt{c-\frac{1}{4}}$

Hence the least distance is

$c\le$if $0\le c\le \frac{1}{2}$ and $\sqrt{c-\frac{1}{4}}$ is $\frac{1}{2}\le c\le 5$