x^2-10xy+y^2+1=0

I'm supposed to rotate the axes in order to eliminate the xy-term. I'm told this problem is very simple, but I don't understand this section at all. Any help is greatly appreciated.

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- Jun 4th 2007, 06:16 PMalreadyinuseRotation and Systems of Quadratic Equations
x^2-10xy+y^2+1=0

I'm supposed to rotate the axes in order to eliminate the xy-term. I'm told this problem is very simple, but I don't understand this section at all. Any help is greatly appreciated. - Jun 4th 2007, 06:37 PMThePerfectHacker
Given a general conic:

$\displaystyle Ax^2+Bxy+Cy^2+Dx+Ey+F=0$

If $\displaystyle B\not = 0$ (that means it has a cross term).

Then the required angle is:

$\displaystyle \cot 2\theta = \frac{A - C}{B}$

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You problem is,

$\displaystyle x^2 - 10xy+y^2 +1 =0$

The required angle satisfies,

$\displaystyle \cot 2\theta = \frac{ 1 -1 }{10} = 0$

Thus,

$\displaystyle 2\theta = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$

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When we rotate by $\displaystyle \theta$ we get the new coordinates:

$\displaystyle x'= x\cos \theta - y\sin \theta$

$\displaystyle y'= x\sin \theta + y\cos \theta $

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In this case $\displaystyle \theta = \frac{\pi}{4}$ thus,

$\displaystyle x' = x\cos \frac{\pi}{4} - y\sin \frac{\pi}{4} = x\frac{\sqrt{2}}{2} - y\frac{\sqrt{2}}{2}$

$\displaystyle y'= x\sin \frac{\pi}{4} + y\cos \frac{\pi}{4} = x\frac{\sqrt{2}}{2} + y \frac{\sqrt{2}}{2}$

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Substitute the new x and y coordinates:

$\displaystyle \left( x\frac{\sqrt{2}}{2} - y\frac{\sqrt{2}}{2} \right)^2 - 10 \left(x\frac{\sqrt{2}}{2} - y\frac{\sqrt{2}}{2}\right)\left( x\frac{\sqrt{2}}{2} + y \frac{\sqrt{2}}{2} \right) + \left( x\frac{\sqrt{2}}{2} + y \frac{\sqrt{2}}{2} \right)^2+1=0$

Expand,

$\displaystyle \left( \frac{1}{2}x^2 - xy + \frac{1}{2}y^2 \right) - 10 \left(\frac{1}{2}x^2 - \frac{1}{2}y^2 \right) + \left( \frac{1}{2}x^2+xy+\frac{1}{2}y^2 \right)+1=0$

Combine,

$\displaystyle -4x^2+6y^2+1=0$ - Jun 4th 2007, 07:14 PMalreadyinuse
Thanks for helping me with the last problem. I know how to solve the rest of my homework except for this one:

xy-2y-4x=0

Am I supposed to change the equation first? - Jun 4th 2007, 07:16 PMThePerfectHacker
- Jun 4th 2007, 07:26 PMalreadyinuse
Ah, okay. *obviously braindead*

Thanks for all your help!