The condition for the line$\mathrm{px} + \mathrm{qy} + \mathrm{r} =$0 may be a normal to the rectangular hyperbola $\mathrm{xy} = {\mathrm{c}}^2$ is

Given Rectangular hyperbola $\mathrm{xy}={\mathrm{c}}^2$ 

let $\mathrm{y}-\frac{\mathrm{c}}{t}={\mathrm{t}}^2(\mathrm{x}-\mathrm{ct})$ be normal to the given hyperbola.

$\Rightarrow {\mathrm{xt}}^3-\mathrm{yt}-{\mathrm{ct}}^4+\mathrm{c}=0 - ①$

Given normal $\mathrm{px}+\mathrm{qy}+\mathrm{r}=0 - ②$

Since $① \& ②$ represents same line then

$$\begin{gathered} \Rightarrow \frac{{\mathrm{t}}^3}{\mathrm{p}}=\frac{-\mathrm{t}}{\mathrm{q}}=\frac{\mathrm{c}-{\mathrm{ct}}^4}{\mathrm{r}} \Rightarrow \frac{{\mathrm{t}}^3}{\mathrm{p}}=\frac{-\mathrm{t}}{\mathrm{q}} \Rightarrow {\mathrm{t}}^2=\frac{-\mathrm{p}}{\mathrm{q}} \\ \Rightarrow \mathrm{p}>0, \mathrm{q}<0 (\therefore {\mathrm{t}}^2 \mathrm{is} \mathrm{positive}) \\ OR \\ Given curve xy=c^2 \\ \Rightarrow y=\frac{c^2}{x} \\ differentiate w.r.to x on both sides \\ \frac{dy}{dx}=\frac{-c^2}{x^2} \\ slope of normal to the given curve=\frac{x^2}{c^2}>0 \\ Given slope of normal px+qy+r=0 is \frac{-p}{q} which is greater than zero if p and q have opposite signs \end{gathered}$$