Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be $\frac{5}{4}$. If the equation of the normal at the point $\left(\frac{8}{\sqrt{5}},\frac{12}{5}\right)$ on the hyperbola is $8\sqrt{5}x+\beta y=\lambda ,$ then $\lambda -\beta$ is equal to

$e^2=1+\frac{b^2}{a^2}=\frac{25}{16}\Rightarrow \frac{b^2}{a^2}=\frac{9}{16}\dots \dots .\left(1\right)$

$A\left(\frac{8}{\sqrt{5}},\frac{12}{5}\right)$ satisfies $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\Rightarrow \frac{64}{5a^2}-\frac{144}{25b^2}=1 .......\left(2\right)$

Solving $\left(1\right) \& \left(2\right) b=\frac{6}{5}, a=\frac{8}{5}$

$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{A}\left(\frac{8}{\sqrt{5}},\frac{12}{5}\right) \mathrm{is} \\ \frac{\sqrt{5}{a}^{2}x}{8}+\frac{5b^{2}y}{12}=a^2+b^2 \\ \frac{\sqrt{5}}{8}\left(\frac{64}{25}\right)x+\frac{5}{12}\left(\frac{36}{25}\right)y=4 \\ \Rightarrow 8\sqrt{5}x+15y=100 \end{gathered}$$

Comparing it with $8\sqrt{5}x+\beta y=\lambda$

Gives $\lambda =100,\beta =15$

$\lambda -\beta =85$