Official solution
Here, tx - 2y - 3t = 0 \& x - 2ty + 3 = 0
On solving, we get;
y = \[{{6t} \over {2{t^2} - 2}}\] = \[{{3t} \over {{t^2} - 1}}\] \& x = \[{{3{t^2} + 3} \over {{t^2} - 1}}\]
Put t = tan\[\theta \]
\[ \therefore \] x = - 3 sec 2\[\theta \] \& 2y = 3 (- tan 2\[\theta \])
\[ \because \] sec22\[\theta \] - tan22\[\theta \] = 1
\[ \Rightarrow \] \[{{{x^2}} \over 9}\] - \[{{{y^2}} \over {9/4}}\] = 1
which represents at hyperbola
\[ \therefore \] a2 = 9 \& b2 = 9/4
\[\lambda (T.A.) = 6; e2 = 1 + \[{{9/4} \over 9}\] = 1 + \[{1 \over 4}\] \[ \Rightarrow \] e = \[{{\sqrt 5 } \over 2}\]