<s>[INST] Shortest distance of the point $(0,c)$ from the parabola $y=x^2$ Shortest distance of the point $(0,c)$ from the parabola $y=x^2$ ? (Where $0\le c \le5$) 
My approach: I wrote the distance formula by taking parametric coordinates as $(t,t^2)$ and then differentiated the equation.I got the extremum as $x=\sqrt (c)$,but that's the wrong answer? I can't figure out my mistake.
Please help! 
 [/INST] Given is
$$
y = x^2
$$
So the normal for point $(x_o,x_o^2)$ can be written as
$$
(x,y) = (x_o - 2\alpha x_o, x_o^2 + \alpha )
$$
The length of the normal from point $(x_o,x_o^2)$ is given by
$$
\ell = \alpha \sqrt{4 x_o^2 + 1}
$$
We need to solve
$$
(0,c) = (x_o - 2\alpha x_o, x_o^2 + \alpha )
$$
or
$$
\left[
\begin{array}{rcl}
x_o - 2 \alpha x_o &=& 0\\
c &=& x_o^2 + \alpha
\end{array}
\right.
$$
From
$$
x_o - 2 \alpha x_o = 0
$$
follows $x_o$ or $\alpha = \tfrac{1}{2}$.
Case $x_o=0$
We obtain
$$
\left[
\begin{array}{rcl}
0 &=& 0\\
c &=& \alpha
\end{array}
\right.
$$
Thus $\alpha = c$ and $x_o=0$. So the length is given by
$$
\ell = c
$$
Case $\alpha = \tfrac{1}{2}$
We obtain
$$
\left[
\begin{array}{rcl}
0 &=& 0\\
c &=& x_o^2 + \frac{1}{2}
\end{array}
\right.
$$
Thus $\alpha = \tfrac{1}{2}$ and $x_o = \sqrt{c - \tfrac{1}{2}}$. The case $c<\tfrac{1}{2}$ givens the $x_o=0$ solution.
The case $c \ge \frac{1}{2}$ gives the length
$$
\ell = \sqrt { c^2 - \tfrac{1}{4} } 
$$
Note that $c \ge \sqrt{ c^2 - \tfrac{1}{4}}$ for $c \ge \frac{1}{2}$, so the shortest distance is given by
$$
c \textrm{ for $c \le \tfrac{1}{2}$}
$$
or
$$
\sqrt{c^2 - \tfrac{1}{4}} \textrm{ for $c \ge \tfrac{1}{2}$}
$$
</s>