Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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im really bored
2006-01-06 12:02:09

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.

krassi_holmz
2006-01-05 20:56:18

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

krassi_holmz
2006-01-05 20:54:36

krassi_holmz
2006-01-05 20:53:04

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.

im really bored
2006-01-05 15:57:32

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????

im really bored
2006-01-05 11:23:25

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?

im really bored
2006-01-05 09:47:02

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

krassi_holmz
2006-01-05 02:02:26

Can be 180 deg.

Ricky
2006-01-05 01:48:53

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

krassi_holmz
2006-01-04 22:22:23

I can't understand, too:

krassi_holmz
2006-01-04 22:14:40

What angle of rotation?

im really bored
2006-01-04 18:35:52

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?