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Problem:

Given the equation $$5x^2 + 5y^2 - 6xy - 8 = 0$$ defining a non-degenerate conic section, find a rotation of the variables, such that the cross term $-6xy$ disappears in the new coordinates $(\bar x, \bar y)$.

Attempts:

Given the standard 2D rotation matrix $\textbf R$, I figured I'd invert it, and use that to find a way to get $\bar x \bar y = 0$.

That's where I'm stuck. I can't seem to find a way to do this, given $\textbf R^{-1} = \begin{bmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$.

Any tips?

Alec
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3 Answers3

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Did you try a less clever method?

Put $x=\alpha x' + \beta y'$ and $y=-\beta x' + \alpha y'$

Then the equation

$5x^2 + 5y^2 - 6xy - 8 = 0$ becomes

$(5(\alpha^2+\beta^2)+6\alpha\beta)x'^2+(5(\alpha^2+\beta^2)-6\alpha\beta)y'^2-6x'y'(\alpha^2-\beta^2)-8=0$

If you want to remove any cross product, then $\alpha^2-\beta^2=0$

EDITED after abel comment (forgotten contributing $x'^2$ and $y'^2$ coming from the $xy$ term).

It is easy to finish from here...

Martigan
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  • I get that you can insert $\alpha^2 + \beta^2$ wherever, since that's just $1$, but how do you justify inserting $\alpha^2 - \beta^2$ in the $xy$ term? Aren't you effectively changing the value of the term? – Alec Jan 13 '15 at 09:24
  • @Aleksander I just changed $x$ by $\alpha x' +\beta y'$ each time there was $x$, and the same for $y$. – Martigan Jan 13 '15 at 09:28
  • now you are back to $\alpha = \cos \theta, \beta = \sin \theta$ what have you gained by not staring with the rotations as the op did. – abel Jan 13 '15 at 10:02
  • @abel True, I don't say it is fundamentally different. I just tried to show a way that would be a bit alternative and perhaps easier to get to the (same) result. – Martigan Jan 13 '15 at 10:33
  • how is that an alternative way? you still have to deal with $\alpha^2 + \beta^1 = 1$ – abel Jan 13 '15 at 10:36
  • @abel Finding $\theta$ is very easy given $\alpha^2-\beta^2=0$, isn't it? And that is the whole point of the exercise, after all... – Martigan Jan 13 '15 at 10:44
  • your answer is not correct as it is now. you forgot the contributions to ${x^\prime}^2$ and ${y^\prime}^2$ from the $xy$ term. – abel Jan 13 '15 at 10:49
  • @abel You are right. I have been a bit quick trying to solve the $x'y'$ issue. – Martigan Jan 13 '15 at 11:08
  • @Martigan - In my attempt, I get $$xy = (\alpha x' - \beta y')(\beta x' + \alpha y') = x'y'(\alpha^2 - \beta^2) - \alpha\beta(x'^2-y'^2)$$

    I can of course cancel the term with $\alpha^2-\beta^2$ with an angle of pi/4, but I'm not sure what to do with the other term. Are you somehow implementing that into the other terms?

     – Alec Jan 13 '15 at 14:55
  • @Aleksander, you need to combine/add the squared terms with other squared terms coming from $5x^2$ and $5y^2.$ – abel Jan 13 '15 at 15:01
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define new variables $$x = x_1\cos t - y_1 \sin t , y = x_1 \sin t + y_1 \cos t$$ substituting in $5(x^2+y^2) - 6xy - 8= 0$ you get $$5(x_1^2 + y_1^2) - 6(x_1\cos t - y_1\sin t)(x_1\sin t + y_1 \cos t) - 8 = 0$$ which simplifies to $$x_1^2(5 - 6 \sin t \cos t) + y_1^2(5 +6\sin t \cos t)- 6x_1y_1(\cos^2 t - \sin ^2 t) - 8 = 0$$

you can eliminate the $x_1y_1$ term if you choose $t = 45^\circ$ which reduces the equation to the ellipse $$x_1^2 + 4y_1^2 = 4.$$

and if you choose $t = 135^\circ$ which reduces the equation to the ellipse $$4x_1^2 + y_1^2 = 4.$$

abel
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  • In your final stage, with $t = 45^o$, we get $\sin t = \cos t = \frac1{\sqrt2}$ which gives $\sin t \cos t = \frac12$ right? Then don't we end up with $8x_1^2 + 2y_1^2 = 8$? – Alec Jan 13 '15 at 16:06
  • @Aleksander, you are right. i will edit it. – abel Jan 13 '15 at 16:07
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Given $$Ax^2+Bx+Cy^2+Dy+Exy+F=0,$$ Then $$tan(2\theta)=\frac{E}{C-A},$$ for $$x=\bar{x}cos(\theta)+\bar{y}sin(\theta),\quad y=-\bar{x}sin(\theta)+\bar{y}cos(\theta).$$

Ofir Schnabel
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  • The only drawback of this formula is the precise case here where $A=C$. – Martigan Jan 13 '15 at 09:31
  • True, but it happens when $cos(2\theta)=0$. Then you know $\theta$. Thank you for pointing that out. – Ofir Schnabel Jan 13 '15 at 09:33
  • I've seen the $\tan(2\theta) = \frac{E}{C-A}$ a few times, but I don't understand where that comes from. – Alec Jan 13 '15 at 09:45
  • Put $x=\bar{x}cos(\theta)+\bar{y}sin(\theta),\quad y=-\bar{x}sin(\theta)+\bar{y}cos(\theta),$ in a general equation and find the conditions such that the cofficient of $\bar{x}\bar{y}$ is zero. – Ofir Schnabel Jan 13 '15 at 09:50
  • If I use $\tan(2\theta) = E/(C-A)$ and I want $E=0$ for the xy term to cancel out, I get $\tan(2\theta) = 0$ so $\theta = \pi/2$. From other answers, I think the answer is supposed to be $\pi/4$ which solves $cos^2\theta-\sin^2\theta = 0$. Am I missing something? – Alec Jan 13 '15 at 15:01
  • I was looking all over for a derivation of this, but this comment encouraged me to go and derive it myself, so kudos. If anyone else is doing the same, the important part is that you can ignore most of the irrelevant variables and just focus on the cross products from A, B and C. And remember your double angle formulae! – Sam Keays Dec 10 '24 at 22:23