# Which surface is formed by rotating a hyperbola around its asymptotes?

I don't know even what a type of surface will be. And what equation will be? The equation of hyperbola - $$xy = l.$$ Now, let's

$$x = x'cos(\varphi ) - y'sin(\varphi ), y = x'sin(\varphi ) + y'cos(\varphi ) \Rightarrow \frac{1}{2}sin(2 \varphi )x'^{2} - \frac{1}{2}sin(2 \varphi )y'^{2} + x'y'cos(2 \varphi) = l.$$

So

$$cos( 2 \varphi ) = 0 \Rightarrow \varphi = \frac{\pi}{4} \Rightarrow \frac{x'^{2}}{2l} - \frac{y'^{2}}{2l} = 1.$$

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A rotated hyperbola!(?) But, what kind of answer do you expect? – draks ... May 14 '12 at 21:29
It's equation. I don't know. – John Taylor May 14 '12 at 21:30
Unfortunately, some object doesn't work. – John Taylor May 14 '12 at 21:44
Do you want to rotate it around $x$ and find the result, then go back to the original hyperbola and rotate it around $y$ and find the result? Or rotate it around $x$ and rotate the resulting surface around $y$? The second result looks like a mess to me, and it will depend upon the order of rotation. – Ross Millikan May 14 '12 at 21:48
other hiperbola, sure... – Gastón Burrull May 15 '12 at 0:27

Hint: If you rotate a curve in the $xy$ plane around the $x$ axis, each point $(a,b,0)$ will trace out a circle parallel to the $yz$ plane. The $x$ position of all points on the circle will be the same as the original point, $a$, all the points on the circle will be the same distance from the axis, $b$.