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Hyperbolic Paraboloid and Isometry 
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#1
Nov712, 11:21 PM

P: 84

If the hyperbolic paraboloid z=(x/a)^2  (y/b)^2
is rotated by an angle of π/4 in the +z direction (according to the right hand rule), the result is the surface z=(1/2)(x^2 + y^2) ((1/a^2)((1/b^2)) + xy((1/a^2)((1/b^2)) and if a= b then this simplifies to z=2/(a^2) (xy) suppose z= x^2  y^2 does this mean that z=2xy ? if so can someone tell me how to put z=x^2  y^2 into it's quadric form? Also the rotation by the angle of π/4 is that just the typical rotation matrix Rz? 


#2
Nov812, 05:48 AM

Sci Advisor
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Thanks
P: 26,148

Hi BrainHurts!
(try using the X^{2} button just above the Reply box ) 


#3
Nov812, 11:45 AM

P: 84

i actually got it,
a quadric is an equation that can be written in the form: v'Av + bv + c = 0 so z=x^2  y^2 by the above equation: let v' is the transpose of v and v = [x y z] A= [ 1 0 0; 0 1 0; 0 0 0] and b = [0 0 1] so rotating v by the rotation matrix where the angle is pi/4 about the z axis gave me the results I was looking for. Thank you very much sorry i don't see the button over the reply box but i'll try to do it next time! 


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