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Eliminate the parameter. x = 3 cos t, y = 3 sin t
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\(\bf x = 3cos(t)\qquad \qquad y = 3sin(t)\\ \quad \\ \quad \\ \textit{let us solve first by the cos(t) and sin(t)}\\ x = 3cos(t)\implies\cfrac{x}{3}=cos(t)\\ \quad \\ y = 3sin(t)\implies \cfrac{y}{3}=sin(t)\\ \quad \\ \textit{recall the trig identity of }\quad sin^2(\theta)+cos^2(\theta) = 1\\ \quad \\ \cfrac{x}{3}=cos(t)\qquad \cfrac{y}{3}=sin(t)\qquad sin^2(t)+cos^2(t) = 1\quad thus\\ \quad \\ \left(\cfrac{x}{3}\right)^2+\left(\cfrac{y}{3}\right)^2=1\) raise both numerator and denominators to their exponent, and you'd end up with an ellipse :)
Thank you very much!!! :D
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