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.
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,
$\displaystyle -4x^2+6y^2+1=0$