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

$a$

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

Then,

$\sqrt{x_1}+\sqrt{y_1}=\sqrt{a}$                  .......(i)

Now,

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

$\Rightarrow \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{\sqrt{y}}{\sqrt{x}} \Rightarrow\left(\frac{d y}{d x}\right)_P=-\sqrt{\frac{y_1}{x_1}}$

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

$y-y_1=-\sqrt{\frac{y_1}{x_1}}\left(x-x_1\right)$

$y-y_1=-\sqrt{\frac{y_1}{x_1}}\left(x-x_1\right)$

$\Rightarrow \frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\sqrt{x_1}+\sqrt{y_1} \Rightarrow \frac{x}{\sqrt{x_1}}+\frac{y}{\sqrt{y_1}}=\sqrt{a}$

[Using (i)]

This cuts the coordinate axes at $A\left(\sqrt{a x_1}, 0\right)$ and $B\left(0, \sqrt{a y_1}\right)$

∴  $O A+O B=\sqrt{a x_1}+\sqrt{a y_1}=\sqrt{a}\left(\sqrt{x_1}+\sqrt{y_1}\right)=\sqrt{a} \times \sqrt{a}=a$