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The arclength of a circle is just:

$$r\theta$$

is there relation like this for a hyperbola? for example:

$$r\phi$$

where phi is the argument of the hyperbolic functions.

Habouz
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1 Answers1

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Using the parametrization $$x=a \cosh(\theta) \qquad \text{and} \qquad y=b\sinh(\theta)$$ you need to compute $$\int \sqrt{a^2 \sinh ^2(\theta )+b^2 \cosh ^2(\theta )}\,d\theta=-i b E\left(i \theta \left|\frac{a^2}{b^2}+1\right.\right)$$ where appears (just as for the ellipse) the elliptic integral of the second kind.

Now, use bounds.