text: Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4
  
  • May 28th 2011, 01:34 PM
  Lizziepaquette
  Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4
  In essence I'm trying to take a horizontal ellipse and put it at an angle, rotating it around the origin. It's for my conics project, I'm making the elliptical orbits of the planets.
  
  So I have the a, b, and c terms of the ellipse, (as in the distance from vertice to center and such) which are 28, 21, and $7\sqrt{7}$ respectively. When plugged into the equation they give:
  
  ${x}^{2} / 784 +{y}^{2} /441 =1$
  
  which turns into
  
  $441 {x}^{2} + 784{y}^{2 } - 345,744 = 0$
  
  from staring at my math book i've assumed that these x's and y's are actually x prime and y prime in the grand scheme of how we normally look at rotations.
  
  so actually
  
  $441{x'}^{2 } + 784{y'}^{2 } - 345,744 = 0$
  
  with this is mind I tried to apply the transformation equations knowing that I was trying to acheive a rotation of $\pi / 4$
  
  $x = x' \cos \pi /4 - y' \sin \pi /4$
  
  $x = (x' - y')/\sqrt{2}$
  
  $\sqrt{2}x = x' - y'$
  
  $\sqrt{2}y = x' + y'$
  
  But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable and create a Bxy term.
  
  So please, its due Wednesday... help?
  • May 29th 2011, 12:20 AM
  Opalg
  Quote:
  
  Originally Posted by Lizziepaquette
  I was trying to achieve a rotation of $\pi / 4$
  
  $x = x' \cos \pi /4 - y' \sin \pi /4$
  
  $x = (x' - y')/\sqrt{2}$
  
  $\sqrt{2}x = x' - y'$
  
  $\sqrt{2}y = x' + y'$
  
  But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable
  
  If you get from $(x',y')$ to $(x,y)$ by rotating through $\pi/4$, then you can get back from $(x,y)$ to $(x',y')$ by rotating through $-\pi/4$. Therefore if $x = x' \cos \pi /4 - y' \sin \pi /4$ then $x' = x \cos(- \pi /4) - y \sin(- \pi /4) = \tfrac1{\sqrt2}x +\tfrac1{\sqrt2}y.$
  • May 29th 2011, 06:47 AM
  Lizziepaquette
  Hmm let me plug it in
  • May 29th 2011, 06:48 AM
  Lizziepaquette
  I did the math out with that concept and my end result is
  
  $597.5{x}^{2} -373xy + 597.5{y}^{2} -345,744 = 0$
  
  I haven't done the forward math to check but i graphed it and it looks about right.
  
  Thanks (Smile)