TITLE: Difference between length of the entire hyperbola and that of its asymptotes
QUESTION [1 upvotes]: In this paper on page 7 it says that the difference between the whole length of the hyperbola and that of its asymptotes, $\Delta$, is given by $$\Delta=4a\int_0^{\alpha}\sqrt{1-e^2\sin^2\theta}d\theta,$$ where $\sin\alpha=1/e$, $e$ is the eccentricity, and this is for a hyperbola with equation $x^2/a^2-y^2/b^2=1$. No proof was given and I couldn't find one online.
I have tried to find expressions for the length of the asymptote and the length of the hyperbola in the 1st quadrant using polar coordinates with $L=\int_{\theta_0}^{\theta_1}\sqrt{r^2+(dr/d\theta)^2}d\theta$, but the difference was always an ugly expression instead of what was given(I'm not sure if converges if I do it this way). I also tried taking $\theta$ to be the acute angle between the tangent to the hyperbola and its asymptote(in the 1st quadrant) but I do not know how to interpret the integral geometrically or otherwise. I also tried substituting $u=\sqrt{1-e^2\sin^2\theta}$, but I cannot make sense of the resulting integral.
I have no idea how to proceed now, can someone please help?

REPLY [2 votes]: Let's take just one quarter of the hyperbola and its asymptotes, namely that part lying in the first quadrant. One could try to give a sense to that difference by integrating the difference of line elements, for instance along $y$.
The line element of the asymptote is $dl_a=\sqrt{1+a^2/b^2}dy$ and that of the hyperbola is $dl_h=\sqrt{1+a^4/b^4\cdot y^2/x^2}dy$. The difference in length could then be computed as
$$
\begin{align}
{\Delta\over4}&=\int_0^{+\infty}\left(
\sqrt{1+{a^2\over b^2}}-\sqrt{1+{a^4\over b^4}{y^2\over x^2}}
\right)dy\\
&=\sqrt{1+{a^2\over b^2}}\int_0^{+\infty}\left(
1-\sqrt{1-{a^2\over a^2+b^2}{b^2\over b^2+y^2}}
\right)dy\\
&=ae\int_0^{+\infty}\left(
1-\sqrt{1-{1\over e^2}{1\over 1+t^2}}
\right)dt,\\
\end{align}
$$
which is indeed convergent (and can be written in terms of elliptic functions).
I don't know how to transform the above integral into the one you found in that paper, but I checked with Mathematica and they give the same result.
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