How does Rotating an Ellipse change its equation?

[resurgance2001]

Homework Statement

For an ellipse x^2/a^2 + y^2/a^2 = 1 prove that is the ellipse is rotated counter clockwise by an angle of 45 degrees the new equation for the ellipse is (x + y)^2/2a^2 + (x - y)^2/a^2 = 1

Relevant Equations

Rotation matrix plus parametric equations for an ellipse.

 

[resurgance2001]

when I apply the same rotation matrix to a simple vector I do get a counter clockwise rotation of 45 degrees. But when I try to do the same thing with the equation for the ellipse the rotation is clockwise. Is the mistake that what I have actually done is to rotate the axis? I am confused about this and the textbook solution is not very helpful. See below:

 

[mathwonk]

Let T be the matrix of rotation counterclockwise by 45 degrees. Then based on what this does to the basic unit vectors, (1,0) and (0,1) the matrix T has columns [1/sqrt(2), 1/sqrt(2)], [-1/sqrt(2), 1/sqrt(2)]. Then if g = 0 is the equation for the original ellipse, and f=0 is the equation for the rotated ellipse, we have g(x) = 0 if and only if f(T(x)) = 0, if and only if (T*f)(x) = 0, where T* is pulling back functions via T. Hence g = T*f, so f = (T*)^-1(g). But pulling back polynomials is determined by the pullback on linear, i.e. coordinate functions, and pulling back linear functions is just done by the transpose matrix. But since T is a rotation matrix, if T* is the transpose matrix, then T* = T^-1, so (T*)^-1 = T. So since in the dual space the function x has coordinates (1,0) and y has coordinates (0,1), they transform by x-->x/sqrt(2) + y/sqrt(2) and y--> -x/sqrt(2) + y/sqrt(2). Thus the equation x^2/a^2 +y^2/b^2 =1, transforms into the equation obtained from these substitutions, i.e. (x+y)^2/2a^2 + (-x+y)^2/2b^2 = 1, where (-x+y)^2 = (x-y)^2.

 

[vela]

You ended up with $x' = (x-y)/\sqrt 2$ and $y' = (x+y)/\sqrt 2$. The line $x'=0$, which is where $x=y$, is the $y'$-axis, and the line $y'=0$, which is where $x=-y$, is the $x'$-axis. When you rotate the ellipse counterclockwise, you want its axis on the $y'$-axis to have length $a$ and length $b$ on the $x'$ axis, but you wrote the equation of the ellipse in terms of $x'$ and $y'$ the other way around.

 

[vela]

Is the mistake that what I have actually done is to rotate the axis?

You made two mistakes. You rotated the axes and then wrote down the equation for a different ellipse than what you started with.

There are two ways to attack the problem. The book's solution is an active rotation. It took points on the ellipse $(a \cos\theta, b\sin\theta)$ and rotated them about the origin by +45 degrees. In other words, they left the axes alone and rotated the ellipse.

The other way would be to leave the ellipse alone and rotate the axes by -45 degrees. This is a passive rotation. If you rotate the x- and y- axes in the clockwise direction by 45 degrees, then relative to these axes, the original ellipse appears rotated by 45 degrees in the counterclockwise direction.

Your attempt rotated the axes by 45 degrees clockwise, but then you wrote down the equation for a different ellipse that aligned with those axes. That ellipse is rotated 45 degree clockwise relative to the original ellipse, which is what you discovered. What you needed to do was solve for x and y in terms of x' and y' and then substitute back into the equation for the ellipse.

 

[resurgance2001]

Thank you. Is there a textbook that you can recommend which covers this topic - specifically in relation to rotating hyperbolic equations?

Thanks again.

 

[resurgance2001]

You made two mistakes. You rotated the axes and then wrote down the equation for a different ellipse than what you started with.

There are two ways to attack the problem. The book's solution is an active rotation. It took points on the ellipse $(a \cos\theta, b\sin\theta)$ and rotated them about the origin by +45 degrees. In other words, they left the axes alone and rotated the ellipse.

The other way would be to leave the ellipse alone and rotate the axes by -45 degrees. This is a passive rotation. If you rotate the x- and y- axes in the clockwise direction by 45 degrees, then relative to these axes, the original ellipse appears rotated by 45 degrees in the counterclockwise direction.

Your attempt rotated the axes by 45 degrees clockwise, but then you wrote down the equation for a different ellipse that aligned with those axes. That ellipse is rotated 45 degree clockwise relative to the original ellipse, which is what you discovered. What you needed to do was solve for x and y in terms of x' and y' and then substitute back into the equation for the ellipse.

Thank you. Can you recommend a textbook which covers specifically the topic of rotated hyperbolic equations? Thanks again.

 

[resurgance2001]

You ended up with $x' = (x-y)/\sqrt 2$ and $y' = (x+y)/\sqrt 2$. The line $x'=0$, which is where $x=y$, is the $y'$-axis, and the line $y'=0$, which is where $x=-y$, is the $x'$-axis. When you rotate the ellipse counterclockwise, you want its axis on the $y'$-axis to have length $a$ and length $b$ on the $x'$ axis, but you wrote the equation of the ellipse in terms of $x'$ and $y'$ the other way around.

Thank you. I would really like to see this in a text book. Can you recommend one please?

 

[vela]

I don't know of a textbook off the top of my head. It's been so long since I first ran across rotations. You could try looking through the OpenStax math textbooks, available for free download. I'd start with the books on college algebra/trig, calculus (look up something like change of coordinates), and linear algebra.

Perhaps someone else will chime in to give you better suggestions.

 

[resurgance2001]

I don't know of a textbook off the top of my head. It's been so long since I first ran across rotations. You could try looking through the OpenStax math textbooks, available for free download. I'd start with the books on college algebra/trig, calculus (look up something like change of coordinates), and linear algebra.

Perhaps someone else will chime in to give you better suggestions.

Thank you. I have now found one online resource which is dedicated to this specific topic which looks very useful. I will also check out the resource you have suggested.

 

[vanhees71]

Obviously they look at an active rotation of the ellipse, counter-clockwise by $\pi/4$. In 2D Cartesian coordinates you have
$$\vec{x}'=\hat{D} \vec{x}, \quad \hat{D}=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 ^1 \end{pmatrix}.$$
What you need is the inverse, which is easy, because $\hat{D}^{-1}=\hat{D}^{\text{T}}$ for an O(2) matrix. So you have
$$\begin{pmatrix}x \\ y \end{pmatrix}=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 ^1 \end{pmatrix} \begin{pmatrix} x' \\ y' \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} x'+y' \\ y'-x' \end{pmatrix}.$$
So you just can plug this in the original equation of the ellipse,
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \; \Rightarrow \; \frac{(x'+y')^2}{2 a^2}+\frac{(x'-y')^2}{2 b^2}=1.$$

 

[resurgance2001]

Thanks - I can see that now. To find the equation of the ellipse in the original axes we actually need to use the inverse of the rotation matrix. Thanks again.

 

[PeroK]

Thank you. I have now found one online resource which is dedicated to this specific topic which looks very useful. I will also check out the resource you have suggested.

The key point you have to be clear about is what you are actually doing. One process is to take a vector, rotate it, and write down the new vector components in terms of the original components. That's not what you are doing here.

The required process is as follows. You have a set of points $(x,y)$ that satisfy an equation. Under a rotation, each of these points transforms to a new point $(x',y')$. As above, you can specify the new coordinates in terms of the original coordinates. But, if you think about it, that's not so useful.

What you really want is the original coordinates expressed in terms of the new coordinates. Because then you can substitute the expressions for $x$ and $y$ into the original equation and get an equation for $x'$ and $y'$.

What you want, therefore, is to write $x = f(x',y')$ and $y = g(x',y')$. This works in general, not just for rotations. That's why it's the inverse transformation that you really want. And, of course, for an anticlockwise rotation the inverse is a clockwise rotation.

 

[vanhees71]

The ellipse equation can be written in components wrt. the Cartesian basis with the semiaxes in direction of the basis vectors as
$$\vec{x}^{\text{T}} \hat{M} \vec{x}=1, \quad \hat{M}=\mathrm{diag}(1/a^2,1/b^2).$$
In an arbitrarily rotated Cartesian basis you have
$$\vec{x}'=\hat{D} \vec{x},$$
where
$$\hat{D}=\begin{pmatrix} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi \end{pmatrix}$$
is the rotation matrix. Then
$$\vec{x}=\hat{D}^{\text{T}} \vec{x}'$$
Plugging this into the equation, gives
$$\vec{x}^{\prime \text{T}} \hat{M}' \vec{x}'=1, \quad \hat{M}'=\hat{D} \hat{M} \hat{D}^{\text{T}}.$$