Answer
$$e = \frac{{3\sqrt 3 }}{2}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can see:}} \cr
& {\text{Focus }}\left( {27,0} \right){\text{ and the point }}\left( { - 27,48\frac{3}{4}} \right),\,\,{\text{directrix }}x = 4 \cr
& {\text{The equation of the hyperbola is }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{With foci }}\left( { \pm c,0} \right),{\text{Focus }}\left( {27,0} \right) \to c = 27 \cr
& \cr
& {\text{The directrix on a hyperbola is given by }}x = \pm \frac{a}{e} \cr
& {\text{directrix }}x = 4 \cr
& 4 = \frac{a}{e} \cr
& {\text{The exccetricity of a hyperbola is }}e = \frac{c}{a} \cr
& 4 = \frac{a}{{c/a}} \cr
& 4c = {a^2} \cr
& {a^2} = \left( 4 \right)\left( {27} \right) \cr
& a = \sqrt {\left( 4 \right)\left( {27} \right)} \cr
& a = 6\sqrt 3 \cr
& \cr
& e = \frac{{27}}{{6\sqrt 3 }} \cr
& e = \frac{{3\sqrt 3 }}{2} \cr} $$
