You are not logged in.
#1 2006-01-04 18:35:52
- im really bored
- Full Member
Offline
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?
#2 2006-01-04 22:14:40
- krassi_holmz
- Real Member
Offline
Re: Rotation of Conics
What angle of rotation?
IPBLE: Increasing Performance By Lowering Expectations.
#3 2006-01-04 22:22:23
- krassi_holmz
- Real Member
Offline
Re: Rotation of Conics
I can't understand, too:
Last edited by krassi_holmz (2006-01-04 22:22:49)
IPBLE: Increasing Performance By Lowering Expectations.
#4 2006-01-05 01:48:53
- Ricky
- Moderator
Offline
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..."
#5 2006-01-05 02:02:26
- krassi_holmz
- Real Member
Offline
Re: Rotation of Conics
Can be 180 deg.
IPBLE: Increasing Performance By Lowering Expectations.
#6 2006-01-05 09:47:02
- im really bored
- Full Member
Offline
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 (2006-01-05 09:48:45)
#7 2006-01-05 11:23:25
- im really bored
- Full Member
Offline
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?
#8 2006-01-05 15:57:32
- im really bored
- Full Member
Offline
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????
#9 2006-01-05 20:53:04
- krassi_holmz
- Real Member
Offline
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.
IPBLE: Increasing Performance By Lowering Expectations.
#10 2006-01-05 20:54:36
- krassi_holmz
- Real Member
Offline
Re: Rotation of Conics
Here's your ellipse(the first problem):
Last edited by krassi_holmz (2006-01-05 20:55:05)
IPBLE: Increasing Performance By Lowering Expectations.
#11 2006-01-05 20:56:18
- krassi_holmz
- Real Member
Offline
Re: Rotation of Conics
Is this works? (plot is streched a little, but the numbers are correct)
IPBLE: Increasing Performance By Lowering Expectations.
#12 2006-01-06 12:02:09
- im really bored
- Full Member
Offline
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.