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$$x=\frac{2t+1}{2t^{2}+2t+1}$$ $$y=\frac{2t^{2}+2t}{2t^{2}+2t+1}$$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?

user71207
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2 Answers2

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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.

José Carlos Santos
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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.

Toby Mak
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  • Sorry, I dont really get this. How did you obtain $y$ in this first place? Any why is showing that the $x$ coordinate is different for one value of $t$ sufficient to prove for all values of $t$? Induction comes to my mind – user71207 Feb 24 '21 at 11:04
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    Write the numerator as $(2t^2 + 2t + 1) - 1$ first. – Toby Mak Feb 24 '21 at 11:04
  • Is $x$ really not odd? I thought an odd function divided by an even function makes an odd function. – user71207 Feb 24 '21 at 11:13
  • The numerator of $x$ is neither odd nor even, and so is the denominator. – Toby Mak Feb 24 '21 at 11:14
  • Where does $ t - 2(t + 0.5) = -t - 1 $ come from? – user71207 Feb 24 '21 at 11:17
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    It helps to imagine $t > -0.5$ (although this applies to any given $t$). $t$ has distance $t - (-0.5) = t + 0.5$ from $-0.5$, and the reflection is the same distance away from $0.5$, hence $t - (t + 0.5) - (t + 0.5) = t - 2(t+0.5)$. – Toby Mak Feb 24 '21 at 11:19
  • Recall that the definition of a one-to-one function is that $f(a) = f(b)$ implies $a = b$. – Toby Mak Feb 24 '21 at 11:26
  • Ok so was there a particular reason you considered $y$ instead of the $x$ parametric? – user71207 Feb 24 '21 at 11:35
  • $y$ is easier to manipulate than $x$. I would highly recommend that you use technology such as Desmos (graphing software) to help with visualising the functions and getting an overall feel for the problem. – Toby Mak Feb 24 '21 at 11:40
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    And no, induction will not help because you are dealing with real numbers, not integers. It's easier to substitute $t$ and $-t-1$ into $x$ as from the graph, those are the only two values of $t$ for which the $y$-coordinates are equal. – Toby Mak Feb 24 '21 at 11:40