My Math Forum parameterization of curve

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 August 10th, 2017, 05:18 AM #1 Member   Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0 parameterization of curve I wonder whether this is correct because later, when trying to find the osculating circle, I get huge numbers and a lot of roots. x^2+y^2-z^2=1 x+y+z=0 Trying to parametrize this, I got: z=-x-y x^2+y^2-(-x-y)^2=1 x^2+y^2-(x^2+2xy+y^2)=1 -2xy=1 xy=-1/2 So: x=t y=-1/2t z=(-2t^2+1)/2t Last edited by skipjack; August 10th, 2017 at 06:22 AM.
 August 10th, 2017, 06:31 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,188 Thanks: 1648 The 2t denominator should be enclosed in parentheses, but your work is otherwise correct. Thanks from sarajoveska

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