Rotate Parabola: Eliminate xy-Term

In summary, the conversation involves finding the angle of rotation to eliminate the xy-term in the given equation. By using the formula cot(2θ)= (A-C)/B and substituting the values of A, B, and C, the correct angle of rotation is found to be approximately 71.57 degrees. However, in the attempt at a solution, the angle was incorrectly calculated as 1/sqrt(3) instead of sqrt(3)/3. This error was corrected and the simplified equation was found to be 2x^2+ 2sqrt(3)xy+ y^2.
  • #1
themadhatter1
140
0

Homework Statement


Rotate the axis to eliminate the xy-term.

[tex]3x^2-2\sqrt{3}xy+y^2+2x+2\sqrt{3}y=0[/tex]

Homework Equations


[tex]\cot2\theta=\frac{A-C}{B}[/tex]
[tex]x=x'\cos\theta-y'\sin\theta[/tex]
[tex]y=x'\sin\theta+y'\cos\theta[/tex]

The Attempt at a Solution



Find the Angle of Rotation

[tex]\cot2\theta=\frac{-1}{3}=\frac{\cot^2\theta-1}{2\cot\theta}[/tex]
[tex]-(3\cot\theta-1)(\cot\theta+3)=0[/tex]
[tex]\cot\theta=\frac{1}{3}[/tex]
[tex]\theta\approx71.57 degrees[/tex]

By drawing a triangle you can find

[tex]\sin\theta=\frac{3}{\sqrt{10}}[/tex]
[tex]\cos\theta=\frac{1}{\sqrt{10}}[/tex]
Then using the 2 equations you find
[tex]x=\frac{x'-3y'}{\sqrt{10}}[/tex]
[tex]y=\frac{3x'+y'}{\sqrt{10}}[/tex]

Substitute that back in and you get

[tex]3(\frac{x'-3y'}{\sqrt{10}})^2-2\sqrt{3}(\frac{x'-3y'}{\sqrt{10}})(\frac{3x'+y'}{\sqrt{10}})+(\frac{3x'+y'}{\sqrt{10}})^2+2\frac{x'-3y'}{\sqrt{10}}+2\sqrt{3}\frac{3x'+y'}{\sqrt{10}}[/tex]

Now I need to simplify...

[tex]\frac{3x'^2-18x'y'+27y'^2}{10}+\frac{-6\sqrt{3}x'^2+18\sqrt{3}x'y'+6\sqrt{3}y'^2}{10}+\frac{9x'^2+6x'y'+y'^2}{10}+\frac{2x'-6y'}{\sqrt{10}}+\frac{6\sqrt{3}x'+2\sqrt{3}y'}{\sqrt{10}}[/tex]

Simplify more...[tex]\frac{(12-6\sqrt{3})x'^2+(-12+18\sqrt{3})x'y'+(28+6\sqrt{3})y'^2}{10}+\frac{(2+6\sqrt{3})x'+(-6+2\sqrt{3})y'}{\sqrt{10}}[/tex]

Phew.. that was a lot of typing... Now I could cross multiply and type all of that out as well but if I did my x'y' terms won't cancel out and since that was the whole purpose of this endeavor I went wrong somewhere. Do you know where I made my mistake?
 
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  • #2
Your angle is wrong. How did you get it? Did not you miss a square root?

ehild
 
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  • #3
I saw how I was wrong before and used the angle [tex]\frac{1}{\sqrt{3}}[/tex] and it worked out fine. However, looking back at it again I can't find why I took the square root.

Because

[tex]3\cot\theta-1=0[/tex]
[tex]\cot\theta=\frac{1}{3}[/tex]
 
  • #4
ehild is right.

The formula you start with (although the LaTex doesn't appear to be right) is
[tex]cot(2\theta)= \frac{A- C}{B}[/tex]

And here, A= 3, C= 1, ande [itex]B= 2\sqrt{3}[/itex]

[tex]cot(2\theta)= \frac{3- 1}{2\sqrt{3}}= \frac{1}{\sqrt{3}}= \frac{\sqrt{3}}{3}[/tex]
not "1/3".

You didn't take the square root- it is already in the equation: [itex]2x^2+ 2\sqrt{3}xy+ y^2[/itex]
 
  • #5
HallsofIvy said:
ehild is right.

The formula you start with (although the LaTex doesn't appear to be right) is
[tex]cot(2\theta)= \frac{A- C}{B}[/tex]

And here, A= 3, C= 1, ande [itex]B= 2\sqrt{3}[/itex]

[tex]cot(2\theta)= \frac{3- 1}{2\sqrt{3}}= \frac{1}{\sqrt{3}}= \frac{\sqrt{3}}{3}[/tex]
not "1/3".

You didn't take the square root- it is already in the equation: [itex]2x^2+ 2\sqrt{3}xy+ y^2[/itex]

Ok, Now I see where I went wrong. Thanks.
 

FAQ: Rotate Parabola: Eliminate xy-Term

1. How do you rotate a parabola and eliminate the xy-term?

To rotate a parabola, you need to use a transformation matrix. The xy-term can be eliminated by choosing the right values for the rotation angle and the coordinates of the vertex.

2. What is the purpose of rotating a parabola and eliminating the xy-term?

Rotating a parabola can help simplify its equation and make it easier to graph or analyze. Eliminating the xy-term can also make the equation easier to work with and understand.

3. Can a parabola be rotated by any angle?

Yes, a parabola can be rotated by any angle. However, the angle of rotation will affect the shape, orientation, and location of the parabola.

4. How do you know if the xy-term has been eliminated after rotating a parabola?

If the xy-term has been eliminated, the equation of the parabola will only contain terms involving x and y squared. The coefficient of the xy-term will also be equal to 0.

5. Are there any real-world applications for rotating a parabola and eliminating the xy-term?

Yes, rotating a parabola and eliminating the xy-term can be useful in various fields such as engineering, physics, and economics. It can help in modeling and analyzing real-world situations, such as projectile motion and cost optimization.

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