a) \(\left\{ \begin{array}{l}x = 2t{\rm{ }}\left( 1 \right)\\y = 3{t^2} - 2{\rm{ }}\left( 2 \right)\end{array} \right.\) \(\left( 1 \right) \Rightarrow t = \dfrac{x}{2}\) thế vào \(\left( 2 \right)\) \( \Rightarrow y = 3.{\left( {\dfrac{x}{2}} \right)^2} - 2 = \dfrac{3}{4}{x^2} - 2\) (PT quỹ đạo) b) \(\left\{ \begin{array}{l}x = {2^{3t}}{\rm{ }}\left( 1 \right)\\y = {2^{ - 3t + 1}}{\rm{ }}\left( 2 \right)\end{array} \right.\) \(y = {2^{ - 3t + 1}} = \dfrac{2}{{{2^{3t}}}} = \dfrac{2}{x}\) c) \(\left\{ \begin{array}{l}x = 2 + 3\sin 2t\\y = 3 + 3cos2t\end{array} \right.\) \(\begin{array}{l}y = 3 + 3cos2t = 3 + 3\sqrt {1 - {{\sin }^2}\left( {2t} \right)} \\ = 3 + 3\sqrt {1 - \dfrac{{{{\left( {x - 2} \right)}^2}}}{9}} \\ = 3 + \sqrt {9 - {{\left( {x - 2} \right)}^2}} \\ = 3 + \sqrt {5 + 4x - {x^2}} \end{array}\)