Find the equation of the tangent to the curve

$\sqrt{x}+\sqrt{y}=a$

Differentiating both sides w.r.t. $x$,

$\Rightarrow \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=\frac{-\sqrt{y}}{\sqrt{x}}$

Given $\left(x_{1}, y_{1}\right)=\left(\frac{a^{2}}{4}, \frac{a^{2}}{4}\right)$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)\left(\frac{a^{2}}{4}, \frac{a^{2}}{4}\right)=\frac{-\sqrt{\frac{a^{2}}{4}}}{\sqrt{\frac{a^{2}}{4}}}=-1$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-\frac{a^{2}}{4}=-1\left(x-\frac{a^{2}}{4}\right)$

$\Rightarrow y-\frac{a^{2}}{4}=-x+\frac{a^{2}}{4}$

$\Rightarrow x+y=\frac{a^{2}}{2}$