If ${\mathrm{L}}_1 = 0$ and ${\mathrm{L}}_2 = 0$ are the asymptotes of the hyperbola $9{\mathrm{x}}^2-4{\mathrm{y}}^2+36\mathrm{x}+8\mathrm{y}-4=0$ then the product of the perprndiculars from the point (1,1) to the lines ${\mathrm{L}}_1 = 0$ and ${\mathrm{L}}_2 = 0$ is 
 

Given hyperbola $\boxed{\mathrm{S}\equiv 9{\mathrm{x}}^2-4{\mathrm{y}}^2+36\mathrm{x}+8\mathrm{y}-4=0}$

Equation of pair of asymptotes is

$$\begin{gathered} {\mathrm{S}}^1\equiv 9{\mathrm{x}}^2-4{\mathrm{y}}^2+36\mathrm{x}+8\mathrm{y}+\mathrm{k}=0 \to \left(1\right) \\ \frac{\partial s}{\partial x}=0 \Rightarrow 18\mathrm{x}+36=0 \Rightarrow \mathrm{x}=-2 \\ \frac{\partial s}{\partial y}=0 \Rightarrow -8\mathrm{y}+8=0 \Rightarrow \mathrm{y}=1 \\ \mathrm{Centre}=(-2, 1) \end{gathered}$$

since $\left(1\right)$ passes through centre $(-2, 1)$ then 

$$\begin{gathered} 9\left(4\right)-4\left(1\right)+36(-2)+8\left(1\right)+\mathrm{k}=0 \\ \Rightarrow 36-4-72+8-\mathrm{k}=0 \\ \Rightarrow \mathrm{k}=32 \\ \therefore {\mathrm{S}}^1\equiv 9{\mathrm{x}}^2-4{\mathrm{y}}^2+36\mathrm{x}+8\mathrm{y}+32=0 \to \left(2\right) \end{gathered}$$

Product of perpendiculars from $(1, 1)$ to $\left(2\right)$

$$\begin{gathered} {\mathrm{d}}_1 {\mathrm{d}}_2=\frac{\left|{\mathrm{S}}_{11}^1\right|}{\sqrt{{(\mathrm{a}-\mathrm{b})}^2+4h^2}} \\ =\frac{\left|9-4+36+8+32\right|}{\sqrt{{\left(13\right)}^2+0}} \\ =\frac{81}{13} \end{gathered}$$