Why is there a point on the unit circle that is not represented by these parametric equations?

Question

$$x=\frac{2t+1}{2t^{2}+2t+1}$$ $$y=\frac{2t^{2}+2t}{2t^{2}+2t+1}$$By squaring $x$ and $y$ and adding them up, I obtained $x^2 + y^2 = 1$ after some algebraic manipulation. But the question asks which point is not represented by these parametric equations. The first thing that came to my mind was to set the denominator does not equal to zero but obviously, it was the wrong this to do. What am I supposed to do in this situation?

Part 2 of the question asks whether the mapping from parameters to points is one to one or many to one. I have identifies that $x$ is odd, so it is definitely one-to-one. However, $y$ is represented by an even function so it cannot by one to one. The answer says overall it is indeed one-to-one - how can this be if $y$ is even? Unless the mapping is only defined by the $x$ function?

Answer 1

It can never be the point $(0,1)$. Note that$$\frac{2t^2+2t}{2t^2+2t+1}=1\iff2t^2+2t=2t^2+2t+1,$$which is impossible.

Answer 2

For part $2$, $x$ is not odd: check your calculations again. It is easier to manipulate $y$ as follows:

$$y = 1 - \frac{1}{2x^2+2x+1} = 1 - \frac{1}{2(x^2+x+0.5)} = 1 - \frac{1}{2(x + 0.5)^2 + 0.5}$$

which is symmetric across $t = -0.5$. So for any given $t$, only $t$ and its reflection $t - 2(t + 0.5) = -t - 1$ have the same $y$-coordinate. If you can show that $x$-coordinates are different for these two values of $t$, this implies one-to-oneness.