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 = $${{\sqrt 5 } \over 2}$$