## Calculus: Early Transcendentals 8th Edition

The sum of the x- and y-intercepts of any tangent line to the curve $~~\sqrt{x}+\sqrt{y} = \sqrt{c}~~$ is equal to $c$
$\sqrt{x}+\sqrt{y} = \sqrt{c}$ 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 $c$