I was solving the following parametric equation: $$r(t)=(x(t);y(t))=(sin^2 (t), cos(t)) t \varepsilon [0; 2 \pi )$$
I thought that I could do this so I could eliminate the parameter:
$$1=sin^2 (t) + cos^2 (t)$$ $$1-cos^2 (t)=sin^2 (t)$$ $$1-y^2 =x$$
The excercise asks for the orientation of the curve.
So if I graph $y(t)$ I get the cosine function. So from 0 to $\pi$ the function dicreases and from $\pi$ to $2 \pi$ the function increases, which means that the particle has two movements. Then, I can graph the $x(y)$ that I got from the Pythagorean identity, which is a quadratic function. Then, putting $t$ values in x and y equations, the first point ($t=0$) tells me that the particle in in $(0,1)$; the second point ($t= \pi $) tells me that it is in $(0;-1)$; the third point is in $t=\pi$, which is equal to the second one; and the fourth point ($t=2 \pi$) tells me that the particle is in $(0;1)$.
Hence, $x(y)$ is just a piece of the whole function, which goes from 1 to -1 in the y axis, and then from -1 to 1.