I have to find the rectangular equation for $x = \dfrac{t+1}{t}$ and $y = \dfrac{t - 1}{t}$.

If I solve for $t$ in terms of $x$, I get $t = \dfrac{1}{x - 1}$, and I substitute this into $y$, and get $y = - x + 2$.

However, the correct answer is $x^2 - y^2 = 4$ and it is solved a different way.

Why is my method wrong?

My Calculations

$x=\frac{t+1}{t}$. And so by rearrangement, $t=\frac{1}{x-1}$

And $y=\frac{t-1}t$, can be simplified by substituting in for $x$. This gives us $y=(1/(x-1)-1)$, which simplifies to $y=(1/(x-1)-1)*((x-1)/1)$

After reducing this expression, I get $y=-x+2$, which is clearly not equal to the correct answer listed above. I am interested in knowing what I have done in my working, and how I can get to the correct answer $x^2-y^2=4$.

  • 2
    $\begingroup$ @jjagmath they did share their calculation. $\endgroup$ May 12 at 2:41
  • $\begingroup$ Taking simple examples, $x=2$ and $y=0$ satisfies all three cases (with $t=1$), but $x=-2$ and $y=0$ only satisfies $x^2-y^2=4$ as there is no satisfactory $t$. $\endgroup$
    – Henry
    May 12 at 14:58

2 Answers 2


find the rectangular equation for $x = \dfrac{t+1}{t}$ and $y = \dfrac{t - 1}{t}$

$\;\dots\,$ the correct answer is $x^2 - y^2 = 4$

$x^2 - y^2 = 4\,$ is the correct answer for the parametrization $\,x=\dfrac{\color{red}{t^2}+1}{t}, y=\dfrac{\color{red}{t^2}-1}{t}\,$, so that's most likely a typo in the problem statement.


Notice that $$x^{2}-y^{2}=\left(\frac{t+1}{t}\right)^{2}-\left(\frac{t-1}{t}\right)^{2}=\frac{4}{t},\quad t\not=0$$

But since solving for $y$ and $x$ we get $$\frac{1}{x-1}=\frac{-1}{y-1}\iff y+x=2$$ So you're right.


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