Parametric coordinates of parabola, ellipse

Question

We know that in parabola, ellipse parametric coordinates is given by $(at^2 , 2at)$ and $(a \cos \theta,b \sin \theta)$ respectively, my query is while deriving it when we put $at^2$ into the normal parabola form of $y^2 = 4ax$ we would get $y^2 = 4a^2 t^2$, how we give a reason to reject the negative root or is it that we are making a choice here as such? Similar thing with ellipse too. Often books do this to avoid that: $y/2a = 2x/y$ but why so we can derive that parametric using this?

Answer 1

We're not really "rejecting the negative root" though. Observe $\sqrt{4a^2t^2} = 2|at|$, so to recover all the values that $2at$ attains we must have $\pm\sqrt{4a^2t^2} = \pm 2|at|$.

That being said, this process is not how you would determine the parameterization of $y^2=4ax$. This graph defines $x$ as a function of $y$ as opposed to the other way around, so really we should be subbing $y(t)=2at$ into the equation and then solving for $x(t)$.

The case of ellipse is harder since we neither have $y$ as a function of $x$ nor $x$ as a function of $y$. Thus when we determine a parameterization for the ellipse it is more sensible to find a common parameterization for the two half-ellipses $y=\pm b\sqrt{1-(x/a)^2}$. Choosing $+$, it is not hard to see that if $x(t)=a\cos(t)$ then $y(t) = b\sqrt{1-\cos^2(t)} = b\sin(t)$ for $t\in [0,\pi]$. We know to take the positive root $\sin(t)$ rather than $-\sin(t)$ since $y\geq 0$ for the half-ellipse $y=b\sqrt{1-(x/a)^2}$ (assuming $b>0$).

Similarly, for half-ellipse $y=-b\sqrt{1-(x/a)^2}$, if we choose $x(t) = a\cos(t)$ for $t\in [\pi,2\pi]$ then $y(t) = -b\sqrt{1-\cos^2(t)} = -b(-\sin(t)) = b\sin(t)$. We know to take $-\sin(t)$ since $\sin(t)\leq 0$ on $t\in[0,\pi]$, so we could not have $y\leq 0$ if we took $+\sin(t)$ as the root. Yet then we've shown $(x(t),y(t)) = (a\cos(t),b\sin(t))$ parameterizes both halves of the ellipse on $[0,\pi]$ and $[\pi,2\pi]$ respectively, so it parameterizes the entire ellipse on $[0,2\pi]$.