Official solution
Given hyperbola, \[{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1\]
Point P (3, 3) is on the parabola
\[ \therefore \] \[{9 \over {{a^2}}} - {9 \over {{b^2}}} = 1\] ...(1)
Equation of normal at (x1, y1),
\[{{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2}{e^2}\]
Normal at p(3, 3),
\[{{{a^2}x} \over 3} - {{{b^2}y} \over 3} = {a^2}{e^2}\] ... (2)
It intersect x-axis at (9, 0),
Putting in equation (2),
\[{{9{a^2}} \over 3} - 0 = {a^2}{e^2}\]
\[ \Rightarrow 3{a^2} = {a^2}{e^2}\]
\[ \Rightarrow {e^2} = 3\]
Also, \[{e^2} = 1 + {{{b^2}} \over {{a^2}}}\]
\[ \Rightarrow 3 = 1 + {{{b^2}} \over {{a^2}}}\]
\[ \Rightarrow {b^2} = 2{a^2}\]
Putting the value of b2 in equation (1),
\[{9 \over {{a^2}}} - {9 \over {2{a^2}}} = 1\]
\[ \Rightarrow {a^2} = {9 \over 2}\]
\[ \therefore \] \[({a^2},{e^2}) = \left( {{9 \over 2},3} \right)\]