Question

Show that 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.

Answer 1

\[\frac{1}{2}\frac{1}{\sqrt{x}} + \frac{1}{2}\frac{1}{\sqrt{y}} y' = 0\]
so the slope of the tgt line at (x0,y0) is
\[y'=-\frac{\sqrt{y_0}}{\sqrt{x_0}}\]
so the eqn of the tgt line(using point slope form) is :
\[y-y_0=-\frac{\sqrt{y_0}}{\sqrt{x_0}}(x-x_0)\]
plug in x=0 we have y-intercept= \[\sqrt{x_0}\sqrt{y_0}+y_0\]
plug in y=0 we have x-intercpt = \[\sqrt{x_0}\sqrt{y_0}+x_0\]
add them tegether we have,
\[x-intercept+ y-intercept= 2\sqrt{x_0}\sqrt{y_0}+x_0+y_0= (\sqrt{x_0}+\sqrt{y_0})^2=c \]
the last equality holds because (x0,y0) is one the curve