Let point $\mathrm{P}\left({\mathrm{x}}_1 {\mathrm{y}}_1\right)$ on the curve $\mathrm{xy}={\mathrm{c}}^2$

$\begin{array}{rcl}\mathrm{xy}& =& {\mathrm{c}}^2 \Rightarrow \mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0 \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\raisebox{1ex}{$-\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.\\ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left({\mathrm{x}}_1 {\mathrm{y}}_1\right)}& =& \frac{-{\mathrm{y}}_1}{{\mathrm{x}}_1}\end{array}$

eq of Tangent $\mathrm{y}-{\mathrm{y}}_1=\frac{-{\mathrm{y}}_1}{{\mathrm{x}}_1}\left(\mathrm{x}-{\mathrm{x}}_1\right)$

$\begin{array}{rcl}{\mathrm{x}}_1\mathrm{y}-{\mathrm{x}}_1 {\mathrm{y}}_1& =& -{\mathrm{xy}}_1+{\mathrm{x}}_1{\mathrm{y}}_1\\ & \Rightarrow & {\mathrm{xy}}_1+{\mathrm{x}}_1\mathrm{y}=2{\mathrm{x}}_1{\mathrm{y}}_1\end{array}$

meet coordinate axes at A, B

$\begin{array}{rcl} & & \mathrm{A}\left(2{\mathrm{x}}_1, 0\right) \mathrm{B}\left(0, 2{\mathrm{y}}_1\right)\\ \frac{\mathrm{PA}}{\mathrm{PB}}& =& \frac{2{\mathrm{x}}_1-{\mathrm{x}}_1}{{\mathrm{x}}_1-0}=\frac{{\mathrm{x}}_1}{{\mathrm{x}}_1}=\frac{1}{1} \Rightarrow \mathrm{AP} : \mathrm{PB}=1 : 1\end{array}$