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Find a suitable rotation that eliminates the mixed term in the equation: $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$.

Now we want to introduce the new coordinates for $x,y$:

$$x = \cos \theta X - \sin \theta Y$$

$$y = \cos \theta Y + \sin \theta X $$

Now it would be easy to just see the $2xy$ term as $$2xy = 2( \cos^2 \theta Y - \sin^2 \theta X )$$

We would then want to find out when: $0 = \cos^2 \theta Y - \sin^2 \theta X \rightarrow \theta = \frac{\pi}{4}$

But this seems a bit too easy for me. How do I know that there will not be new $xy$ terms when I change the base to: $x = \cos \theta X - \sin \theta Y$, $y = \cos \theta Y + \sin \theta X $?

Thank you!

Nicky Hekster
  • 49,281
Lukas Arvidsson
  • 2,376

1 Answers1

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You need to evaluate all second degree terms, $3x^2−2xy+3y^2$.

In this case it will work, as the coefficients of $x^2$ and $y^2$ are equal, so that the terms $2\cos\theta\sin\theta XY$ will cancel out.

In the general case $Ax^2+Bxy+Cy^2$, the coefficient of the $XY$ term is $$2(C-A)\cos\theta\sin\theta+B(\cos^2\theta-\sin^2\theta)=(C-A)\sin2\theta+B\cos2\theta=0,$$ and $$\theta=\frac12\arctan\frac B{A-C}.$$