Explanation

(b)
On differentiating given curve w.r.t. $x$, we get
$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}.\frac{dy}{dx} = 0$
$\Rightarrow \ \ \ \ \ \frac{dy}{dx} = - \frac{\sqrt{y}}{\sqrt{x}}$
The normal is parallel to $x$-axis, if
$- \left( \frac{dx}{dy} \right)_{(x_{1},\ y_{1})} = 0\ \ \ \ \ \ \Rightarrow \ \ \ \ \ \sqrt{\frac{x_{1}}{y_{1}}} = 0\ \ \Rightarrow \ \ \ \ x_{1} = 0$ …(ii)
Since, the point $(x_{1},\ y_{1})$ lies on a curve
$\therefore\ \ \ \ \ \ \sqrt{x_{1}} + \sqrt{y_{1}} = \sqrt{a}$
$\Rightarrow \ \ \ 0 + \sqrt{y_{1}} = \sqrt{a}$ [from Eq. (ii)]
$\Rightarrow \ \ \ \ \ \ \ y_{1} = a$
$\therefore$ Required point is $(0,\ a)$