Ellipse Rotation: Solving with Normal Rotation Matrix

[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 I am 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

 

[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.

 

[Physgeek64]

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.

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

 

[blue_leaf77]

I have converted into quadratic form to get (13, -3sqrt3, -sqrt3, 7)

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

 

[Physgeek64]

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

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

 

[blue_leaf77]

if you multiple by (x,y)T A (x,y) where A is the matrix stated above.

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.

 

[Physgeek64]

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.

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

 

[blue_leaf77]

What about the coefficient of $x^2$?

 

[Physgeek64]

What about the coefficient of $x^2$?

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

 

[Physgeek64]

What about the coefficient of $x^2$?

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

 

[blue_leaf77]

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

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

 

[Physgeek64]

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

algebraic, but in the unrotated co-ordinate system

 

[blue_leaf77]

algebraic, but in the unrotated co-ordinate system

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.