# Parametrization of $x^2-y^2=1600$

While trying to compute the line integral along a path K on a function, I need to parametrize my path K in terms of a single variable, let's say this single variable will be $$t$$. My path is defined by the following ensemble: $$K=\{(x,y)\in(0,\infty)\times[-42,42]|x^2-y^2=1600\}$$ I know how to calculate the line integral, that is not my issue. My problem is to parametrize $$x^2-y^2=1600$$. I tried using the identities: $$\sin^2(t)+\cos^2(t)=1$$ $$\sec^2(t)-\tan^2(t)=1$$ But I did not get anywhere with my parametrization (see below for my poor try into parametrizing). I would welcome any help/hints and if you happen to know some good reading to learn more about parametrization, I am also interested. $$r(t)=1600\sec^2(t)-1600\tan^2(t)=1600$$ for $$x=40\sec(t) \land y=40\tan(t)$$

• Are you familiar with the hyperbolic functions? – N. F. Taussig Jun 17 '19 at 14:35
• Hello! I am a little bit familiar, yes. I know about the cosh^2-sinh^2=1 but in what degree is it different than using sec^2-tan^2=1? – Bilalord Jun 17 '19 at 14:38
• When parametrizing using $\sec$ and $\tan$ it is not particularly clear what $t$ represents geometrically. When parametrizing using $\sinh$ and $\cosh$ however, then $t$ very nicely corresponds to the angle. See this post – JMoravitz Jun 17 '19 at 14:45

I think that using trigonometric function is overcomplicating it in this case. You can let $$y$$ correspond to a parameter $$t$$, then, since $$x$$ is given to be positive, we can say that $$x$$ is the following positive root $$x = \sqrt{1600 + t^2}.$$ Your parameterised curve is subsequently given by: $$\left\{\left(\sqrt{1600 + t^2},t\right): t \in [-42,42]\right\}.$$

Letting $$y = 40\sinh(t)$$ is also an option, in which case the parameterisation is given by $$\left(40 \cosh(t), 40 \sinh(t) \right).$$ This perhaps looks more appealing although finding the correct bounds on $$t$$ now involves inverse hyperbolic functions, which I will leave up to you if you are willing to do it.

• Thank you Pjotr, this was very helpful! I could solve my problem after reading your answer. In fact, I forgot to consider the "simple" way of parametrization... – Bilalord Jun 17 '19 at 16:48

Notice that $$x^2-y^2=(x+y)(x-y)=1600$$, therefore you are dealing with a hyperbola with asymptotes $$x+y=0$$ and $$x-y=0$$ as shown here:

It is natural to try $$x+y=t$$, therefore $$x-y= {1600\over t}.$$ That gives $$x = {t+ {1600\over t} \over 2} \quad y= {t- {1600\over t} \over 2}.$$ As far as bounds go, $$-42\leq y \leq 42$$ therefore $$-84 \leq t- {1600\over t} \leq 84 \rightarrow 16\leq t \leq 100.$$

• Thank you for your answer A.G. I would not have seen this way to parametrize and it is interesting to me to consider your idea! – Bilalord Jun 17 '19 at 16:50
• “Asymptotes,” not “axes.” – amd Jun 17 '19 at 21:18

$$x^2-y^2=1600$$ $$(\frac{x}{40})^2 - (\frac{y}{40})^2=1$$

Let $$x=40\cdot(e^t+e^-t)/2=40\cdot\cosh{t}$$.

Let $$y=40\cdot (e^t-e^{-t})/2=40 \cdot\sinh{t}$$

Use positive values of t if you need positive values of $$x$$.

$$d/dt(\cosh{t})=\sinh{t}$$

$$d/dt(\sinh{t})=\cosh{t}$$

• Thank you for your answer! Helped me to see it with hyperbolic functions. – Bilalord Jun 17 '19 at 16:51