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#1 2006-01-03 19:35:52

im really bored
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Registered: 2005-05-12
Posts: 76

Rotation of Conics

I dont think im really understanding this topic.  For the problem xy - 1 = 0, how do you get the angle of the rotation out of that?

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#2 2006-01-03 23:14:40

krassi_holmz
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Registered: 2005-12-02
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Re: Rotation of Conics

What angle of rotation?


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#3 2006-01-03 23:22:23

krassi_holmz
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Registered: 2005-12-02
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Re: Rotation of Conics

I can't understand, too:

View Image: rot.GIF

Last edited by krassi_holmz (2006-01-03 23:22:49)


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#4 2006-01-04 02:48:53

Ricky
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Registered: 2005-12-04
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Re: Rotation of Conics

You have to give the full problem.  I'm sure the question isn't just, "Find the angle of rotation in xy - 1 = 0."


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2006-01-04 03:02:26

krassi_holmz
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Re: Rotation of Conics

Can be 180 deg.


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#6 2006-01-04 10:47:02

im really bored
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Registered: 2005-05-12
Posts: 76

Re: Rotation of Conics

No the whole problem wasnt that but thats the only part I was stuck on, I know the angle of rotation is 45°.  Its just a bad example.  The roatation of an ellipse shows it better.
The equation given is  7x² - 6√(3)xy + 13y² - 16 = 0.  Its an ellipse that is rotated, but im not sure how to find that degree of rotation

Last edited by im really bored (2006-01-04 10:48:45)

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#7 2006-01-04 12:23:25

im really bored
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Registered: 2005-05-12
Posts: 76

Re: Rotation of Conics

Ok I figured it out.  Just had to plug the a b and c values into the Cot2θ = ( a - c ) / b.  Solving for that im getting θ as 30°.  Then I plug it into x = x'cos 30 - y' sin 30 and y = x' sin 30 + y' cos 30.
I simplified them into x = (√(3)x' - y')/ 2   and y = (x' + √(3)y')/2. Plugged them both back into
7x² - 6√(3)xy + 13y² - 16 = 0. And got this for the standard form (x')²/ 4  +  (y')²/1  =  1. Im hopeing that is right.  How would I go about graphing that, im not sure where to start?

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#8 2006-01-04 16:57:32

im really bored
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Registered: 2005-05-12
Posts: 76

Re: Rotation of Conics

Well I figured out that one, id show what I came up with but I dont have the slightest clue how to show a graph like that on the computer.

New problem now:    2x² - 3xy - 2y² + 10 = 0
a = 2, b = 3, c = -2
plugging it in,  tan2θ = -3/(2 -(-2))    -3/4
using the reference angle from tan-¹ 3/4 I get 2θ = 143.2 so the angle of rotation would be 71.6°
Plugging that into the other forumlas im gonna get a big mess of decimals....
Wonder if there is any way to keep it rationalized????

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#9 2006-01-04 21:53:04

krassi_holmz
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Registered: 2005-12-02
Posts: 1,908

Re: Rotation of Conics

Well done. Only for graphics.

you can use IMPLICIT PLOT. You have 2d euceidin plane with coordinates xOy.
For any point (x,y) if equation EQU(x,y)==0 then plot it, else don't plot it.


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#10 2006-01-04 21:54:36

krassi_holmz
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Registered: 2005-12-02
Posts: 1,908

Re: Rotation of Conics

Here's your ellipse(the first problem):

View Image: ellrot.GIF

Last edited by krassi_holmz (2006-01-04 21:55:05)


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#11 2006-01-04 21:56:18

krassi_holmz
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Registered: 2005-12-02
Posts: 1,908

Re: Rotation of Conics

Is this works? (plot is streched a little, but the numbers are correct)


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#12 2006-01-05 13:02:09

im really bored
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Re: Rotation of Conics

Yeah it looks just like what I got, and much more accurate.  I never liked drawing elipses.  For the second problem I finnaly figured out how to keep it clean, just use the half angle formulas for the cos and sin values.

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