The point on the curve $\sqrt x + \sqrt y = \sqrt a $ , the normal at which is parallel to the x-axis is
A. $\left( {0,0} \right)$
B. $\left( {0,a} \right)$
C. $\left( {a,0} \right)$
D. $\left( {a,a} \right)$

Hint: First we have to differentiate the given curve with respect to x, then we will find the slope of normal. Since normal is parallel to the x axis, the normal will be zero. So by using the normal we find the value of x and by putting the value of x, we can find the value of y and hence we get the point at the normal of the curve which is parallel to the x axis.

Complete step by step solution: 
Given curve is $\sqrt x + \sqrt y = \sqrt a $
Differentiate it with respect to x
$\dfrac{1}{{2\sqrt x }} + \left( {\dfrac{1}{{2\sqrt y }}} \right)y' = 0$
Now on simplifying we get,
$y' = - \sqrt {\dfrac{y}{x}} $
Since we have to find the point at the normal of the curve which is parallel to the x-axis.
First, we will find the slope of normal that is,
Slope of normal $ = - \dfrac{1}{{y'}}$
By putting the value of $y'$ we get,
Slope of normal $ = - \sqrt {\dfrac{x}{y}} $
We denote the slope of normal by m
If normal is parallel to $x$ axis then $m = 0$
That is, $ - \sqrt {\dfrac{x}{y}} = 0$
Now we get, $x = 0$
Put the value of x in the given curve which is $\sqrt x + \sqrt y = \sqrt a $
By putting $x = 0$
We get, $y = a$
$\left( {0,a} \right)$ is the required point at the normal of the curve which is parallel to the x axis.
Thus, Option (B) is correct.

Note: The Slope of normal denoted by m will be found by using the formula that is $m = - \dfrac{1}{{y'}}$ . By using the slope of normal, find out the value of x and then further put the value of x in the given curve to get the value of y. Keep in mind the negative sign while finding the slope of normal. If normal is parallel to the x axis then the slope of normal will be zero.