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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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    other hiperbola, sure...2012-05-15

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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, $b$.

This is presuming the correct reading of the question is the first in my comment.