## Thomas' Calculus 13th Edition

$(x, \sqrt x) = (\frac{1}{m^2}, \frac{1}{m})$
The coordinates of P are $(x, \sqrt x)$. So the line joining P to the origin (0,0) has a slope of $m = \frac{dy}{dx} = \frac{\sqrt x}{x} = \frac{1}{\sqrt x}$ for x > 0. Then $(x, \sqrt x)$ needs to be expressed into m. x = $\frac{1}{m^2}$, because $\frac{1}{m^2}$ = $\frac{1}{{\frac{1}{\sqrt x}^2}}$ = $\frac{1}{\frac{1}{x}}$ = x $\sqrt x$ = $\frac{1}{m}$, because $\frac{1}{m}$ = $\frac{1}{{\frac{1}{\sqrt x}}}$ = $\sqrt x$ Thus, $(x, \sqrt x) = (\frac{1}{m^2}, \frac{1}{m})$