When The curvature is maximum of $x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$ QUESTION

Find Where The Curvature has an extremum                                                                                                                                                                                                                                                                                                                                                                             ?
$$x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$$
MY APPROACH

$$x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}.   .   .   .   .   (1)$$
$$\Rightarrow y^\frac{1}{2}=a^\frac{1}{2}-x^\frac{1}{2}$$
now differntiating both sides we will get:
$$\frac{1}{2}{y^\frac{-1}{2}}\frac{dy}{dx}=(-1)\frac{1}{2}x^\frac{-1}{2}$$
$$\Rightarrow \frac{dy}{dx}=-(\frac{y}{x})^\frac{1}{2}.   .   .   .   .   .(2)$$
Now differentiating again with respect to x again:
$$\frac{d^2y}{dx^2}=-\bigg(\frac{1}{2}(\frac{y}{x})^\frac{-1}{2}.\frac{d}{dx}(\frac{y}{x})\bigg)$$
$$=-\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{d}{dx}(\frac{1}{x})y-\frac{dy}{dx}(\frac{1}{x})\bigg)\Bigg)$$
$$=-\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{-y}{x^2}-\frac{dy}{dx}(\frac{1}{x})\bigg)\Bigg)$$
now put the value of $\frac{dy}{dx}$ :
$$=\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{y}{x^2}-(\frac{y}{x})^\frac{1}{2}\frac{1}{x}\bigg)\Bigg)$$
After simplifying i got :
$$\frac{d^2y}{dx^2}=\frac{1}{2x}\bigg(\frac{y^\frac{1}{2}}{x^2}-1\bigg).   .   .   .   .(3)$$
but from simplification of equation (2) in terms of $a$ wil result :
$$\frac{dy}{dx}=1-(\frac{a}{x})^\frac{1}{2}$$
here i can easily simlify this to get $\frac{d^2y}{dx^2}$ i.e.
$$\Rightarrow\frac{d^2y}{dx^2}=\frac{\sqrt a}{2x\sqrt x}.   .   .   .   .(4)$$
I dont know why i am unable to reduce (3) to (4).May be there exists some calculation error,even thats not my question.
proceeding to find radius of curvature and curvature :
FROM FORMULA
$$\rho=\frac{\bigg(1+(\frac{dy}{dx})^2\bigg)^\frac{3}{2}}{\frac{d^2y}{dx^2}}$$
putting the value of (2) and (4):
$$\Rightarrow \rho=\frac{\bigg(1+\frac{y}{x}\bigg)^\frac{3}{2}2x\sqrt x}{\sqrt a}$$
$$\Rightarrow\rho=\frac{2(x+y)^\frac{3}{2}}{\sqrt a}$$
so curvature is
$$\frac{1}{\rho}=\kappa=\frac{\sqrt a}{2(2x+a-2\sqrt a\sqrt x)^3/2}$$
[notice that i have put y in terms of a nad x]
NOW BEGINS THE PROBLEM

for being extremum
$\frac{d\kappa}{dx} =0 $ and i have to check the sign of $\frac{d^2\kappa}{dx^2}$ :

as you can see in the image :
$$\frac{d\kappa}{dx}=\frac{3\sqrt a(2-\frac{\sqrt a}{\sqrt x})}{4(2x-2\sqrt x\sqrt a+a)^\frac{3}{2}}$$
Now letting this to zero we have :
$$2=\frac{\sqrt a}{\sqrt x}$$
thus i am getting $x=\frac{a}{4}$ as a critical point.
BUT the answer is given as $\frac{\sqrt 2}{a}$
you can even see by inspection $x=\frac{a}{4}$ is not a critical point.

please help and let me know where i have made mistake.
THIS IS MY HUMBLE REQUEST.
 A: Perhaps not a complete answer, but below you will find my solution to the problem, which leads to the same answer as yours, so we are either both wrong or the answer sheet is wrong. 
Parameterize your curve as
$$ x(t) = a\cos^4 t, \quad y(t) = a\sin^4 t. $$
Then
\begin{align}
\dot x(t) &= -4a\cos^3 t \sin t,\\
\ddot x(t) &= -4a(-3\cos^2t\sin^2t + \cos^4 t)\\
&= -4a\cos^2 t(\cos^2t -3\sin^2 t), 
\end{align}
and
\begin{align}
\dot y(t) &= 4a\sin^3 t \cos t,\\
\ddot y(t) &= 4a(3\sin^2t\cos^2t - \sin^4 t)\\
&= 4a\sin^2 t(3\cos^2t -\sin^2 t).
\end{align}
Then
\begin{align}
\dot x \ddot y - \ddot x \dot y
&= -16a^2\cos^3 t \sin^3 t (3\cos^2t -\sin^2 t) + 16a^2\sin^3 t \cos^3 t(\cos^2t -3\sin^2 t)\\
&= 16a^2\cos^3 t \sin^3 t (-4\cos^2 t - 4\sin^2t) = -64 a^2 \cos^3 t \sin^3 t,
\end{align}
and
\begin{align}
(\dot x^2 + \dot y^2 )^{3/2}
&= (16a^2\cos^6t\sin^2t+16a^2\sin^6t\cos^2t)^{3/2}\\
&= 64a^3\cos^3t\sin^3t (\cos^4t+\sin^4t)^{3/2}.
\end{align}
This gives
$$ \kappa(t) = \frac{\dot x \ddot y - \ddot x \dot y}{(\dot x^2 + \dot y^2 )^{3/2}} = - \frac{1}{a(\cos^4t+\sin^4t)^{3/2}}.$$
Now
\begin{align}
\dot\kappa(t) &= \frac{3}{2a}\frac{-4\cos^3t\sin t + 4\sin^3 t\cos t}{(\cos^4t+\sin^4t)^{5/2}}\\
&= \frac{3}{2a}\frac{4\cos t\sin t(\sin^2t- \cos^2 t}{(\cos^4t+\sin^4t)^{5/2}}\\
&= \frac{3}{2a}\frac{-2\sin(2t)\cos(2t)}{(\cos^4t+\sin^4t)^{5/2}}\\
&= -\frac{3\sin(4t)}{2a(\cos^4t+\sin^4t)^{5/2}}.
\end{align}
So $\dot \kappa = 0$ whenever $\sin(4t) = 0,$ which happens for
$$ 4t = \pi n \Leftrightarrow t = \frac{\pi}{4}n, \quad n\in \mathbb{Z}. $$
Since (I presume that) $x,y > 0$, we are considering $t \in (0, \pi/2)$, so the only valid option is $t = \pi/4$, and then
$$ x\left(\frac{\pi}{4}\right) = a\cos^4\left(\frac{\pi}{4}\right) = \frac{a}{\sqrt{2}^4} = \frac{a}{4}, $$
which confirms your answer.
Also, the graph does seem to indicate that the curvature has an extremum at this point. Note that we are not looking for an extremum of the curve, but rather the extremum of its curvature.
A: the equation $$\sqrt{x}+\sqrt{y}=\sqrt{a}$$ is equivalent to $$y=x-2\sqrt{ax}+a,~~~0\leq x \leq a.$$
Hence, $$y'=1-\frac{a}{\sqrt{ax}},~~~y''=\frac{a}{2x\sqrt{ax}}.$$
Therefore, $$k=\frac{|y'|}{(1+y''^2)^{3/2}}=\frac{1}{2}\sqrt{\frac{a}{(2x-2\sqrt{ax}+a)^3}}.$$
Notice that $$2x-2\sqrt{ax}+a=2\left(\sqrt{x}-\frac{\sqrt{a}}{2}\right)^2+\frac{a}{2}\geq \frac{a}{2}$$ with the equality holding if and only if $\sqrt{x}=\dfrac{\sqrt{a}}{2}$, namely $x=\dfrac{a}{4}$. As a result, $k$ takes its maximum value $k=\dfrac{\sqrt{2}}{a}$ at $x=\dfrac{a}{4}.$
