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 August 6th, 2012, 09:57 AM #1 Member   Joined: Jun 2012 Posts: 76 Thanks: 0 rotation of axes question x^2 + 4xy + y^2 - 3 = 0 rewrite the equation in a rotated x'y' system without ans x'y' term. express the equation involving x' and y' in the standard form of a conic section. my solution is x'^2 - y'^2/3 = 1 would anyone like to check this for me ?
 August 6th, 2012, 12:16 PM #2 Global Moderator   Joined: May 2007 Posts: 6,786 Thanks: 708 Re: rotation of axes question It would help if you would supply the equations transforming (x,y) to (x',y').
 August 6th, 2012, 12:35 PM #3 Senior Member   Joined: Jun 2011 Posts: 164 Thanks: 0 Re: rotation of axes question first you want to find the angle you are rotating, using the formula 1/2(Tan^(-1))(B/(A-C)) is A does not equal to C, however, in this case, they do equal to each other, meaning the angle of the rotation is 45 degrees. let ? = the angle of rotation A'=Acos^2(?)+Bsin?cos?+Csin^2(?) B'=0 C'=Asin^2(?)-Bsin?cos?+Ccos^2(?) D'=0 E'=0 F'=-3 D' and E' are both = 0 because D and E are 0 so A'=1/2+4x(1/2)+1/2 C'=1/2-4x(1/2)+1/2 F'=-3 solve each equation A'=3 C'=-1 F'=-3 Finally, the equation i have is: 3x'^2-y'^2-3=0 if we divide everything by 3 (i believe that is what you did) x'^2-(y'^2)/3-1=0, Which is exactly what you got, i believe you are correct
August 6th, 2012, 01:22 PM   #4
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Re: rotation of axes question

Quote:
 Originally Posted by mathman It would help if you would supply the equations transforming (x,y) to (x',y').
huh? i gave you the problem as it is writen.

Quote:
 Originally Posted by wuzhe first you want to find the angle you are rotating, using the formula 1/2(Tan^(-1))(B/(A-C)) is A does not equal to C, however, in this case, they do equal to each other, meaning the angle of the rotation is 45 degrees. let ? = the angle of rotation A'=Acos^2(?)+Bsin?cos?+Csin^2(?) B'=0 C'=Asin^2(?)-Bsin?cos?+Ccos^2(?) D'=0 E'=0 F'=-3 D' and E' are both = 0 because D and E are 0 so A'=1/2+4x(1/2)+1/2 C'=1/2-4x(1/2)+1/2 F'=-3 solve each equation A'=3 C'=-1 F'=-3 Finally, the equation i have is: 3x'^2-y'^2-3=0 if we divide everything by 3 (i believe that is what you did) x'^2-(y'^2)/3-1=0, Which is exactly what you got, i believe you are correct
thank you! i wouldve typed out my work but its so much haha.

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# x^2-4xy-2y^2=3 rotate conic

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