<s>[INST] find shortest distance from origin to parabolic form suppose that,we have given   following parabolic  arc
$\sqrt{x}+\sqrt{y}=\sqrt{a}$
we are trying to find shortest  distance from origin  to  this line, i think that if  we  rewrite it as $\sqrt{y}=\sqrt{a}-\sqrt{x}$, then
$y=a-2\sqrt{ax}-x$ or if we  rearrange  terms,we get  $y+a-\sqrt{ax}-x=0$
distance from origin $(0,0)$ to  line $Ax+by+c=0$  is equal
  $D=(A\cdot0+b\cdot0+c)/(\sqrt{A^2+b^2})$ so  in my case what   would be minimum distance?  please help me
 [/INST] One asks that the vector $(x,y)$ is orthogonal to the tangent at $(x,y)$ of the curve. Thus, assume that $(x,y)$ and $(x+h,y+k)$ are on the curve and that $h,k\to0$. Then 
$$
\sqrt{x}+\sqrt{y}=\sqrt{a}=\sqrt{x+h}+\sqrt{y+k}=\sqrt{x}\sqrt{1+h/x}+\sqrt{y}\sqrt{1+k/y}
$$
hence
$$
\sqrt{x+h}+\sqrt{y+k}=\sqrt{x}+h/(2\sqrt{x})+o(h)+\sqrt{y}+k/(2\sqrt{y})+o(k).
$$
In particular, $h/\sqrt{x}\sim-k/\sqrt{y}$ hence the tangent vector is $(\sqrt{x},-\sqrt{y})$.
The tangent is orthogonal to $(x,y)$ if and only if $x\sqrt{x}-y\sqrt{y}=0$, that is, $x=y$, that is $x=y=a/4$. The shortest distance is $\sqrt{x^2+y^2}=\sqrt2a/4$.
</s>