# Rotation of an ellipse

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1. Jan 13, 2016

### Physgeek64

Hello! Okay- This is a relatively simple problem, but for some reason I'm having huge difficulty with it.

So I have the equation of an ellipse, x^2-6sqrt3 * xy + 7y^2 =16, which I have converted into quadratic form to get (13, -3sqrt3, -sqrt3, 7) and I need to rotate it using the normal rotation matrix in two dimensions (cos, -sin, cos, sin)

But Im struggling to actually apply the rotation matrix- Do apply it to the quadratic form of the matrix? And if so how can I extract the equation of the rotated ellipse from this?

Any help would be GREATLY appreciated!! Thank you

2. Jan 13, 2016

### blue_leaf77

Assume the equation of ellipse you have there to be written in the already rotated coordinate system $(x',y')$, thus
$$x'^2-6\sqrt{3} x'y' + 7y'^2 =16$$
To obtain the expression of this same ellipse in the unrotated coordinate system, you have to apply the clockwise rotation matrix to the point $(x',y')$. By the way the correct rotation matrix is
$$\left( \begin{array}{cc} \cos \theta & \sin\theta \\ -\sin\theta & \cos \theta \\ \end{array} \right)$$
where $\theta$ is positive for counterclockwise rotation.

Last edited: Jan 14, 2016
3. Jan 13, 2016

### Physgeek64

Okay, so do I simply apply the rotation matrix to the quadratic form?

4. Jan 13, 2016

### blue_leaf77

First of all, how did you get that quadratic form, that doesn't seem to lead to the original ellipse equation.

5. Jan 13, 2016

### Physgeek64

if you multiple by (x,y)T A (x,y) where A is the matrix stated above. Sorry I don't know how to use matrix notation on here- but the T represents transpose

6. Jan 14, 2016

### blue_leaf77

Well yeah I know that, but if you really carry out the matrix operation
$$\left( \begin{array}{cc} x & y \\ \end{array} \right) % \left( \begin{array}{cc} 13 & -3\sqrt{3} \\ -\sqrt{3} &7 \\ \end{array} \right) % \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{cc} x & y \\ \end{array} \right) % \left( \begin{array}{c} 13x-3\sqrt{3}y \\ -\sqrt{3}x + 7y \\ \end{array} \right) = 13x^2 - 4\sqrt{3}xy + 7y^2$$
It cannot be the same as the ellipse equation you have in your first post for whatever constant value in the right hand side.

7. Jan 14, 2016

### Physgeek64

Sorry- that was a typo on my part- both of the non-leading diagonal entries should be -3sqrt3

8. Jan 14, 2016

### blue_leaf77

What about the coefficient of $x^2$?

9. Jan 14, 2016

### Physgeek64

Also a typo- sorry
it is meant to be 13x^2

10. Jan 14, 2016

### Physgeek64

I unfortunately don't know how to edit my original post to correct these though

11. Jan 14, 2016

### blue_leaf77

What are you asked to do? Are you asked to express the rotated ellipse in quadratic form or in algebraic form?

12. Jan 14, 2016

### Physgeek64

algebraic, but in the unrotated co-ordinate system

13. Jan 14, 2016

### blue_leaf77

Then you don't need to take a detour by going to the quadratic form since you already have the algebraic form of the original ellipse. So as I said earlier, imagine you have the ellipse in the already rotated coordinate system $(x',y')$, which means
$$13x'^2-6\sqrt{3} x'y' + 7y'^2 =16$$
At the same time, you also have the matrix relation between $(x',y')^T$ and $(x,y)^T$, related by coordinate rotation. From this, you should be able to obtain expressions for $x'$ and $y'$ in terms of $x$ and $y$ and substitute to the ellipse equation.