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

<p>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1\left( {e = {5 \over 4}} \right)$$</p> <p>So, ${b^2} = {a^2}\left( {{{25} \over {16}} - 1} \right) \Rightarrow b = {3 \over 4}a$</p> <p>Also $\left( {{8 \over {\sqrt 5 }},{{12} \over 5}} \right)$ lies on the given hyperbola</p> <p>So, $${{64} \over {5{a^2}}} - {{144} \over {25\left( {{{9{a^2}} \over {16}}} \right)}} = 1 \Rightarrow a = {8 \over 5}$$ and $b = {6 \over 5}$</p> <p>Equation of normal</p> <p>$${{64} \over {25}}\left( {{x \over {{8 \over {\sqrt 5 }}}}} \right) + {{36} \over {25}}\left( {{y \over {{{12} \over 5}}}} \right) = 4$$</p> <p>$\Rightarrow {8 \over {5\sqrt 5 }}x + {3 \over 5}y = 4$</p> <p>$\Rightarrow 8\sqrt 5 x + 15y = 100$</p> <p>So, $\beta$ = 15 and $\lambda$ = 100</p> <p>Gives $\lambda$ $-$ $\beta$ = 85</p>