Rotating axes to elimnate xy

  • Thread starter TonyC
  • Start date
  • #1
86
0
I am having trouble rotating the axes to eliminate the xy-term.
8x^2+64xy+8y^2+12x+12y+9=0

I know Ax^2+Bxy+Cy^2+Dx+Ey+F=0
however, the professor is looking for an equation not an answer.

Here are my choices and I am stumped:
40x^2 + 12 sq rt2 x + 12 sq rt 2 y + 9 = 0

40x^2 - 24y^2 +12 sq rt2y +9 = 0

-24y^2 + 12 sq rt2x + 12 sq rt 2y + 9 = 0

40x^2 - 24y^ + 12 sq rt2x + 9 = 0

I am not sure what the process is to get to this point. I can run the basic Rotation Theorem for Conics and answer the problem with an angle degree.

PLEASE HELP!
 

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
You can also reduce these conics to their standard Euclidean form using lineair algebra, by diagonalizing the corresponding matrix.
 
  • #3
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
Introduce new variables as follows:
[tex]u=x\cos\theta+y\sin\theta, v=-x\sin\theta+y\cos\theta[/tex]
Here, [tex]\theta[/tex] is the angle between the positive x-axis and the positive u-axis.

Solve for x and y in terms of u and v, and gain:
[tex]x=u\cos\theta-v\sin\theta, y=u\sin\theta+v\cos\theta[/tex]

Substitute these expressions for x and y in your equation, and eliminate the uv-term by making a smart choice of [tex]\theta[/tex]


then, redefine "u" and "v" to be new "x" and "y".
(I don't see much point in doing so, but it seems your professor wants you to do that)
 
Last edited:
  • #4
86
0
Can you show me how?
 
  • #5
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
Put the expressions in u and v for x into the x's place, and similar for the expression for y.

For example, [tex]x^{2}=(u\cos\theta-v\sin\theta)^{2}=u^{2}\cos^{2}\theta-uv\sin(2\theta)+v^{2}\sin^{2}\theta[/tex]
 
  • #6
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
If you haven't done so already, we also have:
[tex]xy=(u\cos\theta-v\sin\theta)(u\sin\theta+v\cos\theta)=\frac{u^{2}-v^{2}}{2}\sin(2\theta)+uv\cos(2\theta)[/tex]
[tex]y^{2}=(u\sin\theta+v\cos\theta)^{2}=u^{2}\sin^{2}\theta+uv\sin(2\theta)+v^{2}\cos^{2}\theta[/tex]
Use these expressions to your heart's content..
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,833
956
You were given [tex]x=u\cos\theta-v\sin\theta, y=u\sin\theta+v\cos\theta[/tex]
Go ahead and replace x and y in your equation by those, do the algebra and then pick θ so that the "uv" term has 0 coefficient.
 
  • #8
saltydog
Science Advisor
Homework Helper
1,582
3
Tony, what happen with all this? You figured the angle via:

[tex]Cot[2\alpha]=\frac{A-C}{B}[/tex]

so you get:

[tex]\alpha=\frac{\pi}{4}[/tex]

Great. Make the substitution:

[tex]x=\overline{x}Cos[\alpha]-\overline{y}Sin[\alpha][/tex]

[tex]y=\overline{x}Sin[\alpha]+\overline{y}Cos[\alpha][/tex]

Substitute those expressions into the equation right. For example the [itex]8x^2[/itex] would be:

[tex]8\left[\frac{\overline{x}}{\sqrt{2}}-\frac{\overline{y}}{\sqrt{2}}\right]^2[/tex]

You can do the rest. Simplify to:

[tex]40\overline{x}^2-24\overline{y}^2+12\sqrt{2}\overline{x}+9=0[/tex]

The [itex]\overline{x}-\overline{y} [/itex] axes are just 45 degrees tilted from the x-y axes. The equation is a hyperbola.

Right?

Edit: Corrected angle formula: it's Cot(2a)
 
Last edited:
  • #9
86
0
Thank you very much. This distance learning is extremely difficult. I am a visual person and your help is truly appreciated.
 

Related Threads on Rotating axes to elimnate xy

Replies
2
Views
3K
Replies
0
Views
2K
Replies
4
Views
584
  • Last Post
2
Replies
27
Views
3K
Replies
9
Views
3K
Replies
8
Views
1K
Replies
4
Views
9K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
963
  • Last Post
Replies
1
Views
1K
Top