Official solution
5x = 4\[\sqrt 5 \]
\[ \Rightarrow \] \[x = {4 \over {\sqrt 5 }}\]
Equation of hyperbola
\[{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1\] it passes through \[\left( {4, - 2\sqrt 3 } \right)\]
\[ \therefore {e^2} = 1 + {{{b^2}} \over {{a^2}}}\]
\[ \Rightarrow {a^2}{e^2} - {a^2} = {b^2}\]
\[ \Rightarrow {{16} \over {{a^2}}} - {{12} \over {{a^2}{e^2} - {a^2}}} = 1\]
\[ \Rightarrow {4 \over {{a^2}}}\left[ {{4 \over 1} - {3 \over {{e^2} - 1}}} \right] = 1\]
\[ \Rightarrow 4{e^2} - 4 - 3 = ({e^2} - 1)\left( {{{{a^2}} \over 4}} \right)\]
\[ \Rightarrow 4(4{e^3} - 7) = ({e^2} - 1){\left( {{{4e} \over {\sqrt 5 }}} \right)^2}\]
\[ \Rightarrow 4{e^4} - 24{e^2} + 35 = 0\]