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Conic section equation

  • Thread starter astrololo
  • Start date
  • #26
200
3
In polar coordinates, (1, 3/2) is ##(\sqrt{13}/2, \theta)## where ##\theta = \arctan(1.5) \approx .98## (radians). If you rotate this by ##\pi/4## you get ##(\sqrt{13}/2, \theta + \pi/4)##, again in polar form. Can you work that out?
Ok I was finally able to draw this cursed parabola. May it go to hell. thank you again for your help! This page helped me a lot : https://en.wikipedia.org/wiki/Rotation_of_axes
 
  • #27
200
3
In polar coordinates, (1, 3/2) is ##(\sqrt{13}/2, \theta)## where ##\theta = \arctan(1.5) \approx .98## (radians). If you rotate this by ##\pi/4## you get ##(\sqrt{13}/2, \theta + \pi/4)##, again in polar form. Can you work that out?
ONE VERY LAST QUESTION. Is my equation in this form : y=(x^2)/2+1 in canonical form right ?
 
  • #28
33,179
4,859
ONE VERY LAST QUESTION. Is my equation in this form : y=(x^2)/2+1 in canonical form right ?
I don't know what "canonical form" means.
 

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