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?
Parametric coordinates of parabola, ellipse
conic sections
Related Solutions
Minimize for $t$
$$d^2(t)=(at^2-p)^2+(2at)^2.$$
$$(d^2(t))'=2(at^2-p)2at+2(2at)2a=0.$$
Then $$t=0\lor\left(p\ge2a\land at^2=p-2a\right),$$
The first case gives
$$d^2=p^2,$$ and the second $$d^2=4a(p-a).$$
Take the smallest of the two.
Suppose your parabola is given by $$y=a x^2+b x + c$$ Its derivative is $y'=2 a x + b$. A vector perpendicular to your parabola at the point $(x,y)$ is thus $$(-2 a x - b,1)$$ Let us normalize this vector $$\vec{n}=\frac{(-2 a x - b,1)}{\sqrt{(2 a x + b)^2+1}}$$ The curves at distance $d$ from your parabola (green in your figure) are thus $$(x,a x^2+b x + c)+d\vec{n}$$ The above formula transforms cartesian-like coordinates $(x,d)$ into the transformed space $$T(x,d)=(x,a x^2+b x + c)+d\vec{n}$$ All we need to do is feed the ellipse expression to this transformation $$T(r_1\cos(\theta)+r_1,r_2\sin(\theta)+r_2)$$ this gives a curve with the desired characteristics.
Example:
This curve touches the straight lines at their midpoints. It also touces the top and bottom curve once, but not at their exact midpoint. If you want a curve that touches the bottom curves at its precise midpoint, you need to parametrize your parabola by its arc length, which is not possible in terms of elementary functions.
Best Answer
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 $[0,2\pi]$.