Answer
$\dfrac{y^2}{9}- \dfrac{x^2}{4} = 1$
Work Step by Step
$x = 2 \tan{t} \hspace{30pt} y = 3 \sec{t}$
$\dfrac{3x}{2} = 3 \tan{t} \hspace{30pt} y^2 = 9 \sec^2 {t}$
$ \dfrac{9x^2}{4} = 9 \tan^2{t}$
$\because y^2 = 9 \sec^2 {t} \hspace{30pt} \therefore y^2 = 9+9 \tan^2 {t} $
$y^2 = 9 +\dfrac{9}{4}x^2 $
$\dfrac{y^2}{9}- \dfrac{x^2}{4} = 1$
