What Is the Parallel of Least Radius on a Hyperboloid of Revolution?

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Homework Statement


Show that on of the hyperboloid of revolution x^2+y^2-z^2=1, the parallel of least radius is the line of striction, ...


What's the parallel of least radius?
 
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The "parallel of least radius" is the line, parallel to an axis, that is shortest from one point on the hyperboloid to another.

Now, clear up my confusion: what is a "line of striction"?
 
Given a ruled surface x(t,v)=a(t)+vw(t), a line of striction is a curve b(t) such that <b'(t),w'(t)>=0 for all t and b lies on the trace of x, ie b(t)=a(t)+u(t)w(t) for some real valued function u(t). It be can then shown that u(t) is given by

u=-\frac{&lt;a&#039;,w&#039;&gt;}{&lt;w&#039;,w&gt;}.

The points of a line of striction are the "central points" of the ruled surface.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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