(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Rotate the axis to eliminate the xy-term.

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

2. Relevant 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]

3. 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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# Homework Help: Rotating a Parabola

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