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

$$\begin{gathered} Given hyperbola \frac{x^2}{16}-\frac{y^2}{9}=1 where a^2=16,b^2=9 \\ Now e=\sqrt{\frac{16+9}{16}}=\frac{5}{4} \\ foci are=(\pm 5,0) \end{gathered}$$

Since (5,0) satisfies the equation of the line $3x+(5-4\sqrt{2})y=15$, so the reflected must pass through (−5,0) and

P is the point of intersection of curve and the ray

So, $P(4\sqrt{2},3)$.
 Equation  of S'P is 

$$\begin{gathered} y-0=\frac{3}{4\sqrt{2}+5}(x+5) \\ \Rightarrow 3x-(4\sqrt{2}+5)y+15=0 \end{gathered}$$