This is a partial hyperbola. You can see that all point on this curve satisfy the hyperbola equation if you elimate the variable 't'.

There are 4 possibilities here:2. x^2=(sint)^2

y^2=(cost)^2

x=sin t

y=cos t

x=-sin t

y=-cos t

x=-sin t

y=cos t

x=sin t

y=-cos t

All are the same curve (except some can the orientation of the closed curve).

Which is a circle since,

x^2+y^2=(sin t)^2+(cos t)^2=1 [Note this is not a correct proof, because I only shown that the point lie on a circle rather than all points are on a circle].

Substitute,3. x=sint

y=(sint)^2

y=x*x=x^2

A parabola

See that x^2+y^2=14. x=(1-t^2)/(1+t^2)

y=(2t)/(1+t^2)

So all points lie on a circle, again this does not show that all the points are the circle (which in fact is untrue). But since t>0 you see that 'y' only have positive terms so it seems to be a positive circle, that is, a semicircle.