If a ray of light incident along the line $3\mathrm{x}+(5-4\sqrt{2})\mathrm{y}=15$ gets reflected from the hyperbola $\frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1,$ then reflected ray goes along the line

$$\begin{gathered} \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1 \\ \Rightarrow {\mathrm{a}}^2=16, {\mathrm{b}}^2=9 \\ \mathrm{e}=\sqrt{\frac{16+9}{16}}=\frac{5}{4} \\ \mathrm{Since} (5, 0) \mathrm{satisfies} \mathrm{the} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{line} 3\mathrm{x}+(5-4\sqrt{2})\mathrm{y}=15 \\ \mathrm{so}, \mathrm{the} \mathrm{reflected} ray \mathrm{pass}es \mathrm{through} (-5, 0) \\ \mathrm{P} \mathrm{is} \mathrm{the} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{curre} \mathrm{and} \mathrm{the} \mathrm{ray} \\ \therefore \mathrm{P}=(4\sqrt{2}, 3) \\ \therefore Eq\mathrm{uation} \mathrm{of} {\mathrm{S}}^1\mathrm{P} \mathrm{is} \mathrm{y}=\frac{3}{4\sqrt{2}+5}(\mathrm{x}+5) \\ \Rightarrow (4\sqrt{2}+5)\mathrm{y}=3(\mathrm{x}+5) \\ \Rightarrow 3\mathrm{x}-(4\sqrt{2}+5)\mathrm{y}+15=0 \end{gathered}$$