The sum of the intercepts made on the axes of coordinates by any tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ is equal to 

Let $P\left(x_1,y_1\right)$ be a point on the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$

Then,

$\sqrt{x_1}+\sqrt{y_1}=\sqrt{a}$

Now, 

$\begin{array}{l}\sqrt{x}+\sqrt{y}=\sqrt{a}\\ \Rightarrow \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}\Rightarrow {\left(\frac{dy}{dx}\right)}_p=-\sqrt{\frac{y_1}{x_1}}\end{array}$

The equation of the tangent to the given curve at point  $P\left(x_1,y_1\right)$ is

$\begin{array}{l}y-y_1=-\sqrt{\frac{y_1}{x_1}}\left(x-x_1\right)\\ \Rightarrow \frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\sqrt{x_1}+\sqrt{y_1}\Rightarrow \frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\sqrt{a}\end{array}$

Using (i)]
This cuts the coordinate axes at $A\left(\sqrt{a{x}_1},0\right)\text{ and }B\left(0,\sqrt{a{y}_1}\right)$

$\therefore OA+OB=\sqrt{a{x}_1}+\sqrt{a{y}_1}=\sqrt{a}\left(\sqrt{x_1}+\sqrt{y_1}\right)=\sqrt{a}\times \sqrt{a}=a.$