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Hyperbolic Paraboloid and Isometry

 
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Nov7-12, 11:21 PM   #1
 

Hyperbolic Paraboloid and Isometry


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?
 
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Nov8-12, 05:48 AM   #2
 
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Hi BrainHurts!

(try using the X2 button just above the Reply box )
Quote by BrainHurts View Post
suppose z= x^2 - y^2

does this mean that z=2xy ?
yes, x2 + y2 = 2* (x - y)/√2 * (x + y)/√2 = 2(Rx)(Ry)
if so can someone tell me how to put z=x^2 - y^2 into it's quadric form?
not following you
Also the rotation by the angle of π/4 is that just the typical rotation matrix Rz?
yes
 
Nov8-12, 11:45 AM   #3
 
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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