Rotate Parabola: Eliminate xy-Term

[themadhatter1]

Homework Statement

Rotate the axis to eliminate the xy-term.

$$3x^2-2\sqrt{3}xy+y^2+2x+2\sqrt{3}y=0$$

Homework Equations

$$\cot2\theta=\frac{A-C}{B}$$
$$x=x'\cos\theta-y'\sin\theta$$
$$y=x'\sin\theta+y'\cos\theta$$

The Attempt at a Solution

Find the Angle of Rotation

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

By drawing a triangle you can find

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

Substitute that back in and you get

$$3(\frac{x'-3y'}{\sqrt{10}})^2-2\sqrt{3}(\frac{x'-3y'}{\sqrt{10}})(\frac{3x'+y'}{\sqrt{10}})+(\frac{3x'+y'}{\sqrt{10}})^2+2\frac{x'-3y'}{\sqrt{10}}+2\sqrt{3}\frac{3x'+y'}{\sqrt{10}}$$

Now I need to simplify...

$$\frac{3x'^2-18x'y'+27y'^2}{10}+\frac{-6\sqrt{3}x'^2+18\sqrt{3}x'y'+6\sqrt{3}y'^2}{10}+\frac{9x'^2+6x'y'+y'^2}{10}+\frac{2x'-6y'}{\sqrt{10}}+\frac{6\sqrt{3}x'+2\sqrt{3}y'}{\sqrt{10}}$$

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

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?

 

[ehild]

Your angle is wrong. How did you get it? Did not you miss a square root?

ehild

 

[themadhatter1]

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

Because

$$3\cot\theta-1=0$$
$$\cot\theta=\frac{1}{3}$$

 

[HallsofIvy]

ehild is right.

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

And here, A= 3, C= 1, ande $B= 2\sqrt{3}$

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

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

 

[themadhatter1]

ehild is right.

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

And here, A= 3, C= 1, ande $B= 2\sqrt{3}$

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

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

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