# Find a rectangular equation from a parametric equation. Why is my approach wrong?

Question

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 $$x^2-y^2=4$$.

• @jjagmath they did share their calculation. May 12 at 2:41
• 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 $t$. May 12 at 14:58

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.