If the normal at $({\mathrm{ct}}_1, \mathrm{c}/{\mathrm{t}}_1)$ on the hyperbola $\mathrm{xy} = {\mathrm{c}}^2$ cuts the hyperbola again at $({\mathrm{ct}}_2, \mathrm{c}/{\mathrm{t}}_2)$, then ${\mathrm{t}}_1{}^3\mathrm{t}_2=$
 

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

let $\mathrm{P}=\left({\mathrm{ct}}_1, \frac{\mathrm{c}}{{\mathrm{t}}_1}\right)$ be any point on this hyperbola.

Now equation of normal at 'P' is $\mathrm{y}-\frac{\mathrm{c}}{{\mathrm{t}}_1}={{\mathrm{t}}_1}^2(\mathrm{x}-{\mathrm{ct}}_1)$

$\Rightarrow {\mathrm{yt}}_1-{{\mathrm{xt}}_1}^3-\mathrm{c}+{{\mathrm{ct}}_1}^4=0$

since it passes through $\left({\mathrm{ct}}_2, \frac{\mathrm{c}}{{\mathrm{t}}_2}\right)$ then

$$\begin{gathered} \Rightarrow \left(\frac{\mathrm{c}}{{\mathrm{t}}_2}\right) {\mathrm{t}}_1-\left({\mathrm{ct}}_2\right) {{\mathrm{t}}_1}^3-\mathrm{c}+{{\mathrm{ct}}_1}^4=0 \\ \Rightarrow {{\mathrm{ct}}_1}^3 ({\mathrm{t}}_1-{\mathrm{t}}_2)-\mathrm{c}\left(1-\frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}\right)=0 \\ \Rightarrow {{\mathrm{ct}}_1}^3({\mathrm{t}}_1-{\mathrm{t}}_2)+\frac{\mathrm{c}({\mathrm{t}}_1-{\mathrm{t}}_2)}{{\mathrm{t}}_2}=0 \\ \Rightarrow ({\mathrm{t}}_1-{\mathrm{t}}_2) \left({{\mathrm{ct}}_1}^3+\frac{\mathrm{c}}{{\mathrm{t}}_2}\right)=0 \\ \Rightarrow {{\mathrm{ct}}_1}^3+\frac{\mathrm{c}}{{\mathrm{t}}_2}=0 (\because {\mathrm{t}}_1\ne {\mathrm{t}}_2) \\ \Rightarrow {{\mathrm{ct}}_1}^3=\frac{-\mathrm{c}}{{\mathrm{t}}_2} \\ \Rightarrow {{\mathrm{t}}_1}^3 {\mathrm{t}}_2=-1. \end{gathered}$$