Let the eccentricity of the hyperbola x ^ 2 a ^ 2 - y ^ 2 b ^ 2 = 1...

Using the formula of eccentricity we get,

${e}^{2} = 1 + \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{25}{16} ⇒ \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{9}{16} . . .$\left(1\right)$</p><p>$$A\left(\dfrac{8}{\sqrt{5}}, \dfrac{12}{5}\right)$ satisfies $\dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{{b}^{2}} = 1$

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

Solving \left(1\right) \text{ and } \left(2\right)$b = \dfrac{6}{5}a = \dfrac{8}{5}$

Normal at $A$ is $\dfrac{\sqrt{5}{a}^{2}x}{8} + \dfrac{5{b}^{2}y}{12} = {a}^{2} + {b}^{2}$

Comparing it $8\sqrt{5}x + \betay = \lambda$

Gives $\lambda = 100, \beta = 15$

So, $\lambda - \beta = 85$