Answer
The sum of the x- and y-intercepts of any tangent line to the curve $~~\sqrt{x}+\sqrt{y} = \sqrt{c}~~$ is equal to $c$
Work Step by Step
$\sqrt{x}+\sqrt{y} = \sqrt{c}$
We can find $y'$:
$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}~y' = 0$
$y' = -\frac{\sqrt{y}}{\sqrt{x}}$
At the point $(x_0,y_0)$, the slope of the tangent line is $-\frac{\sqrt{y_0}}{\sqrt{x_0}}$
We can find the equation of the tangent line at the point $(x_0, y_0)$:
$y-y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}}~(x-x_0)$
When $y=0$:
$y-y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}}~(x-x_0)$
$0-y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}}~(x-x_0)$
$\sqrt{x_0~y_0} = x-x_0$
$x = \sqrt{x_0~y_0} +x_0$
The x-intercept is: $~~\sqrt{x_0~y_0} +x_0$
When $x=0$:
$y-y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}}~(x-x_0)$
$y-y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}}~(0-x_0)$
$y-y_0 = \sqrt{x_0~y_0}$
$y = \sqrt{x_0~y_0} +y_0$
The y-intercept is: $~~\sqrt{x_0~y_0} +y_0$
We can find the sum of the two intercepts:
$\sqrt{x_0~y_0} +x_0+\sqrt{x_0~y_0} +y_0$
$= x_0+2\sqrt{x_0~y_0} +y_0$
$= (\sqrt{x_0} +\sqrt{y_0})^2$
$= (\sqrt{c})^2$
$= c$
The sum of the x- and y-intercepts of any tangent line to the curve $~~\sqrt{x}+\sqrt{y} = \sqrt{c}~~$ is equal to $c$