I'm pretty sure that we can trace a conic section like a circle exactly using a NURBS curve. The first quadrant of a unit circle, for example, can be traced by:
$C(t)=\frac{(1-t)^{2}P_{0}+2t(1-t)P_{1}+2t^{2}P_{2}}{(1-t)^{2}+2t(1-t)+2t^{2}}$
where $P_{0}=\left(0,1\right)$, $P_{1}=\left(1,1\right)$, and $P_{1}=\left(1,0\right)$.
But how do I prove that the curve $x^{2} + y^{2} = 1 : 0 \leq x \leq 1$ is equivalent to $C(t)$?
Your equation for $C(t)$ can be broken down into equations for $x$ and $y$ separately: $$ x(t) = \frac{1-t^2}{1+t^2} \quad ; \quad y(t) = \frac{2t}{1+t^2} $$ It’s easy to check that $[x(t)]^2 + [y(t)]^2 = 1$ for all $t$. This means that every point $C(t)= (x(t),y(t))$ lies on the unit circle. Also it’s clear that $0 \le x(t) \le 1$ if $0 \le t \le 1$. Can you take it from there?
The same sort of reasoning will work whenever you have parametric equations and an implicit equation for a conic. In fact, it will work whenever you have parametric equations and an implicit equation for any curve.
A rational quadratic curve will never quite cover an entire conic — there will always be at least one point missing. For example, your parametric equation $C(t)$ will never give you the point $(-1,0)$ on the unit circle no matter what parameter value $t$ you use.