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

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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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

What angle of rotation?

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I can't understand, too:

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

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Can be 180 deg.

IPBLE: Increasing Performance By Lowering Expectations.

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

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

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

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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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Here's your ellipse(the first problem):

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

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Is this works? (plot is streched a little, but the numbers are correct)

IPBLE: Increasing Performance By Lowering Expectations.

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

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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