Answer
$$e = \frac{3}{2}$$
Work Step by Step
$$\eqalign{
& {\text{Focus }}\left( {9,0} \right){\text{ and the point }}\left( {9, - 7.5} \right),\,\,{\text{directrix }}x = 4 \cr
& {\text{The equation of the hyperbola is }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& {\text{Foci}}\left( { \pm c,0} \right) \to c = 9 \cr
& {c^2} = {a^2} - {b^2} \cr
& {a^2} - {b^2} = 81 \cr
& {b^2} = {a^2} - 81 \cr
& \cr
& {\text{Substitute the point }}\left( {9, - 7.5} \right){\text{ into }}\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{81}}{{{a^2}}} + \frac{{225}}{{4{b^2}}} = 1 \cr
& {\text{Substitute }}{b^2} = {a^2} - 81 \cr
& \frac{{81}}{{{a^2}}} + \frac{{225}}{{4\left( {{a^2} - 81} \right)}} = 1 \cr
& 324{a^2} - 26244 + 225{a^2} = 4{a^4} - 324{a^2} \cr
& 4{a^4} - 873{a^2} + 26244 = 0 \cr
& \left( {4{a^2} - 729} \right)\left( {{a^2} - 36} \right) = 0 \cr
& {a^2} = 36{\text{ and }}{a^2} = 182.25\,,\,\,\,\,{a^2} < {c^2},{\text{ then}}\,\, \cr
& {a^2} = 36 \cr
& a = 6 \cr
& \cr
& {\text{Find the eccentricity}} \cr
& e = \frac{9}{6} \cr
& e = \frac{3}{2} \cr} $$
