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.

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 $x=\dfrac{a}{4}.$
