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This equation is actually the solution to the intersection of the two surfaces $z = x^2 - y^2$, and $z = x^2 + xy - 1$.

I am to parametrise the solution curve, which is noted in the title. A first attempt is below:

\begin{align} z &= x^2 - y^2 = x^2 + xy - 1 &\\ \implies x^2 - y^2 &= x^2 + xy - 1 &\\ -y^2 &= xy - 1 &\\ xy &= y^2 - 1 \end{align}

Now, to parametrise, i.e. $x = f(t)$, $y = f(t)$: \begin{align} xy &= y^2 - 1 &\\ x &= \frac{y^2 - 1}{y} &\\ &= y - \frac{1}{y} \end{align}

Let $y = t \in \mathbb{R}\backslash0$, and hence we have: \begin{cases} x = t - \frac{1}{t} \\[2ex] y = t \end{cases}

Will that suffice? I think not; there must be something more.

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    $\begingroup$ Why do you think there's something more? With $t \neq 0$, all of your steps are reversible to give you your original equation. $\endgroup$ Feb 7, 2019 at 4:47
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    $\begingroup$ You might want to check, as @JohnOmielan seems to suggest, that indeed the parametrization you found covers all points on the curve. As it stands, your argument only shows that all points of the form $(t-\frac{1}{t},t)$ are on the curve. $\endgroup$
    – Servaes
    Feb 7, 2019 at 5:00
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    $\begingroup$ @Servaes, ahh, I get it. I'd have to add in a term for z in terms of t, too, because right now, I've lost the information for z... $\endgroup$
    – SRSR333
    Feb 7, 2019 at 5:13
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    $\begingroup$ You also seem to be missing a sign somewhere; the intersection points satisfy $xy=1-y^2$. $\endgroup$
    – Servaes
    Feb 7, 2019 at 5:22
  • $\begingroup$ @Servaes You are correct. The line $-y^2 = xy - 1$ goes to $xy = y^2 - 1$ instead of $xy = -y^2 + 1$, or equivalently $xy = 1 - y^2$ as you state. $\endgroup$ Feb 7, 2019 at 6:48

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