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I'm given a hyperbolic segment, similar to the parabolic segment shown here:

I know the height of the segment ("h" in the wolfram article), and the length of the line segment joining the endpoints of the hyperbola ("2a" in the wolfram article).

Is it possible to find the area of the segment? Also, does there exist an approximation formula, or rapidly converging method to determine the approximate arc length of the given hyperbola?

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You mean you know the distance between the end points, and you want to know the area between the segment and the Y-axis? Or do you know the difference in X-coordinates of the end points, and you want to know the area of a region bounded (somehow) by the segment and the X-axis? – Beta Jul 26 '12 at 1:18
@Beta: sorry, width was a poor choice of words. yes I know the distance, and I want to find the area of the region bounded by the hyperbolic segment, and the line segment joining the endpoints. – cdk Jul 26 '12 at 1:20
My intuition tells me it is possible. Let's see... – Beta Jul 26 '12 at 1:24
Wait... Do you know the parameters of the hyperbola? Because if you don't, I think I can prove it's impossible. And by "height of the segment", do you mean Y-difference between the endpoints, or Y-value of one of, say, the upper one? – Beta Jul 26 '12 at 1:31
@Beta, how can it be insoluble if the MathWorld article linked above gives the solution! I'm assuming, as does the article, that the figure is symmetric about the "$h$ axis", so to speak. – Rahul Jul 26 '12 at 2:40
up vote 2 down vote accepted

It is not possible to find an answer just from the information supplied.

Consider the hyperbola with equation $$\frac{(y-h)^2}{d^2} -\frac{x^2}{c^2}=1.$$ One branch of this has shape roughly similar to the parabola illustrated in your picture. In particular, it has "height" $h$.

In order for the $x$-intercepts to be at $\pm a$ as in the picture, the relevant condition is $c\sqrt{h^2-d^2}=da$. There are infinitely many hyperbolas for specified $h$ and $a$. The areas are not all the same for these hyperbolas, and neither are the arclengths.

Once the hyperbola is completely specified, arclength, though somewhat unpleasant, can be handled by setting up the usual integral. It is one of the relatively rare cases where the integration can be carried out explicitly in terms of elementary functions.

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translate everything so that the vertices of the hyperbola are on the x axis.

find the partial integral of the positive part of the hyperbola from vertex to the positive intersection, and subtract the integral of the positive part of the line (from the x-intercept to the same intersection point).

do the same thing with the negative parts, and add the absolute value of both parts.

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Did you read the question? – Beta Jul 26 '12 at 1:33
@beta i get that a lot. how hard is it to find the functions of lines and parabolas of which you know multiple parameters? – user35945 Jul 26 '12 at 1:36
A hyperbola and parabola are two different things, bub. – J. M. Jul 26 '12 at 1:36

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