From any point (h, k) four normals can be drawn to the hyperbola ${\mathrm{x}}^2–{\mathrm{y}}^2 = 8$ and if the four feet of normals are $\left({\mathrm{x}}_{\mathrm{r}},y_{\mathrm{r}}\right)\mathrm{r}=1,2,3,4$ then ${\mathrm{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3{\mathrm{x}}_4=$
 

$$\begin{gathered} Given hyperbola x^2-y^2=8 which is eqivalent to xy=4 \\ Let (2t_i,\frac{2}{t_i}) i=1,2,3,4 be points on hyperbola \\ Equation of normal at ( 2t_i,\frac{2}{t_i}) is y-\frac{2}{t_i}={t_i}^2(x-2t_i) \\ \Rightarrow y{t}_i-2={t^3}_i(x-2t_i) \\ \Rightarrow 2{t^4}_i-x{t^3}_i+y{t}_i-2=0 \\ \Rightarrow t_1{t}_2{t}_3{t}_4=-1 \\ Now{x}_1{x}_2{x}_3{x}_4 =-16 \end{gathered}$$