The sum of intercepts on co-ordinate axes made by tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{a}$ is:

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

$\begin{array}{r}\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0,\\ \therefore \frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}\end{array}$

Hence tangent at $(x, y)$ is $Y-y=-\frac{\sqrt{y}}{\sqrt{x}}(X-x)$

or $X\sqrt{y}+Y\sqrt{x}=\sqrt{xy}(\sqrt{x}+\sqrt{y})=\sqrt{axy}$

or $\frac{X}{\sqrt{a}\sqrt{x}}+\frac{Y}{\sqrt{a}\sqrt{y}}=1$

Clearly its intercepts on the axes are $\sqrt{a}\sqrt{x}$ and $\sqrt{a}\sqrt{y}$

Sum of the intercepts $=\sqrt{a}(\sqrt{x}+\sqrt{y})=\sqrt{a}\cdot \sqrt{a}=a.$