Currently studying for finals and came along this question.
Eliminate the parameter $t$ to find the $x-y$ equation of
the curve along which the particle moves.
x = 3cos(t)
y = 2sin(t)
What I have done so far:
Knowing $\sin^2(t) + cos^2(t)=1$, I wanted to give each a value. $$\frac{x}{3} = \frac{3\cos (t)}{3}$$ $$\implies\cos(t)= \frac{x}{3}$$
$$\frac{y}{2} = \frac{2\sin (t)}{2}$$ $$\implies\sin(t)= \frac{y}{2}$$
I then input the values of both $\sin(t)$ and $\cos(t)$ into the equation, so
$$\Big(\frac{y}{2}\Big)^2+\Big(\frac{x}{3}\Big)^2=1$$ $$\implies\frac{y^2}{4}+\frac{x^2}{9}=1$$
$\frac{y^2}{4}+\frac{x^2}{9}=1$ is what I got as my final answer, is that right? Trying to make sure I got the hang of parametric equations.
In the equation of a trajectory it is recomended to solve or for $x$ or for $y$. Then, the final equation should be $y=\sqrt {4-\frac{4x^2}{9}}\ \Leftrightarrow y= 2\sqrt {1-\frac{x^2}{9}} \ $ or $x=\sqrt {9-\frac{9y^2}{4}}\ \Leftrightarrow x= 3\sqrt {1-\frac{y^2}{4}}$
But it depends on what you are asked to do. So your resolution is also correct and it gives us, in this case, a better understanding of the trajectory : it's the graph of an elipse.