Problem:
Solution:
The first equation does not come with squares, so maybe it is easier to start with the first one.
$ \begin{eqnarray*} x &=& \frac{t}{1+t} \\ x + tx &=& t \\ x &=& t - tx \\ &=& t(1 - x) \\ t &=& \frac{x}{1 - x} \end{eqnarray*} $
Now we can substitute $ t $ into $ y $ to eliminate $ t $.
$ \begin{eqnarray*} y &=& 1 - \frac{1}{t^2} \\ &=& 1 - \frac{1}{(\frac{x}{1 - x})^2} \\ &=& 1 - (\frac{1 - x}{x})^2 \\ &=& 1 - (\frac{1}{x} - 1)^2 \\ &=& 1 - (\frac{1}{x^2} - \frac{2}{x} + 1) \\ &=& \frac{2}{x} - \frac{1}{x^2} \\ x^2y &=& 2x - 1 \end{eqnarray*} $
For part (b), first of all note that $ (1, 1) $ is in the variety. Note that by the equation $ t = \frac{x}{1 - x} $, we can substitute any $ x $ we want into the equation, except of course we cannot set $ x = 1 $. So for any point in the variety, we can find the corresponding $ t $, therefore we know the parametric representation does cover all points but $ (1, 1) $ on the variety.
Solution:
The first equation does not come with squares, so maybe it is easier to start with the first one.
$ \begin{eqnarray*} x &=& \frac{t}{1+t} \\ x + tx &=& t \\ x &=& t - tx \\ &=& t(1 - x) \\ t &=& \frac{x}{1 - x} \end{eqnarray*} $
Now we can substitute $ t $ into $ y $ to eliminate $ t $.
$ \begin{eqnarray*} y &=& 1 - \frac{1}{t^2} \\ &=& 1 - \frac{1}{(\frac{x}{1 - x})^2} \\ &=& 1 - (\frac{1 - x}{x})^2 \\ &=& 1 - (\frac{1}{x} - 1)^2 \\ &=& 1 - (\frac{1}{x^2} - \frac{2}{x} + 1) \\ &=& \frac{2}{x} - \frac{1}{x^2} \\ x^2y &=& 2x - 1 \end{eqnarray*} $
For part (b), first of all note that $ (1, 1) $ is in the variety. Note that by the equation $ t = \frac{x}{1 - x} $, we can substitute any $ x $ we want into the equation, except of course we cannot set $ x = 1 $. So for any point in the variety, we can find the corresponding $ t $, therefore we know the parametric representation does cover all points but $ (1, 1) $ on the variety.
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