How to find a trajectory $\sigma(t)$ to represent the ellipse $\{(x,y):4x^{2}+9y^{2}=36\}$? Given the ellipse $\{(x,y):4x^{2}+9y^{2}=36\}$, find a trajectory $\sigma(t)$ which represent it.
So far, I have this:
The standard equation for an ellipse is:
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
So,
$4x^{2}+9y^{2}=36\}$
$=\left \langle \text{Arithmetic: Divide by}\ 36 \right \rangle$
$\frac{4x^{2}}{36}+\frac{9y^{2}}{36}=\frac{36}{36}$
$=\left \langle \text{Arithmetic} \right \rangle$
$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
I don't know how to continue developing the solution, but I think the trajectory could be represented as $(a\cos(t),b\sin(t))$
Some ideas, suggestions to solve this?

For a circle $$x^2+y^2 = r^2$$
we can try
$$x=r\cos(t)$$
$$y=r\sin(t)$$
because
$$x^2+y^2 = r^2\cos^2(t)+r^2\sin^2(t) = r^2$$
So for an ellipse
$$a^2x^2+b^2y^2 = c^2$$
we can try
$$x=\frac{c}{a}\cos(t)$$
$$y=\frac{c}{b}\sin(t)$$
In our case we have
$$2^2x^2+3^2y^2 = 6^2$$
so let us try try
$$x=\frac{6}{2}\cos(t)$$
$$y=\frac{6}{3}\sin(t)$$
Trying:
$$2^2x^2+3^2y^2 = 2^2(\frac{6}{2}\cos(t))^2+3^2(\frac{6}{3}\sin(t))^2$$
$$ = 6^2\cos^2(t) + 6^2\sin^2(t) = 6^2 = 36$$
