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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