ellipse boundary after rotation

Question

Assume I have this vertical ellipse with a certain major axis $a$ and minor axis $b$.

Not rotated

If we take the center of the ellipse to be at $(0,0)$, then the top right small red circle will be at $(b,a)$.

Then I rotate it (say by an arbitrary angle $\theta$) about its center:

Rotated

My question is this: what is the new position of the top right small red circle in this new image after rotation relative to the fixed center? For example at $\theta=90^\circ$ its position will be $(a,b)$.

you mean ($\sup_{x\in E_\theta} x_1,\sup_{x\in E_\theta} x_2)$, where $E_\theta= R_\theta (E)$? — Pocho la pantera, Oct 12 '13 at 09:15
Yes! Is there any mathematical expression saying exactly how that point is related to $a$, $q$ (axis ratio) and $\theta$? For example: at $\theta=90$degrees that point will be on (a,b). — astroboy, Oct 12 '13 at 09:21

score 2 · Accepted Answer · edited Oct 12 '13 at 09:34

2

$$ r(t)=(a\,\cos (t), b\, \sin(t)) $$
After rotation, $$ r_2(t)=R_\theta.r(t)= (a\,cos(t)\cos(\theta)+b\sin(t)\sin(\theta),-a\,cos(t)\sin(\theta)+b\sin(t)\cos(\theta)) $$ So you need to find the maximum of $ a\,cos(t)\cos(\theta)+b\sin(t)\sin(\theta)$ and $-a\,cos(t)\sin(\theta)+b\sin(t)\cos(\theta)$.

Can you do it?

edited Oct 12 '13 at 09:34

astroboy

209

answered Oct 12 '13 at 09:28

Pocho la pantera

2,436

I don't understand what $t$ is? – astroboy Oct 12 '13 at 09:31
$t\in[0,2\pi]$ and $r(t)$ is a parametrization of the ellipse. – Pocho la pantera Oct 12 '13 at 09:33
Do you have some knowledge of calculus? – Pocho la pantera Oct 12 '13 at 09:35
So there is no exact (analytic) answer to this other than testing the whole range of $0\leq t\leq2\pi$? Fortunately I know how to find the maximum of this function ;-). Thanks for the prompt answer. – astroboy Oct 12 '13 at 09:45
You are welcome. There is an exact analytic solution but is bit complicated. – Pocho la pantera Oct 12 '13 at 10:07
I tried finding the analytic answer, thanks again for the explanation. – astroboy Oct 12 '13 at 11:50

astroboy · Answer 2 · 2013-10-12T12:40:14.050

Following Pocho la pantera's explanation, I just thought to extend it. We know that for the rotated ellipse:

$$ x=acos(t)cos(\theta)+bsin(t)sin(\theta) $$

Taking the derivative and setting it to zero we find the $t$ that gives the extremuems of x:

$$ \frac{dx}{dt}=-asin(t)cos(\theta)+bcos(t)sin(\theta)=0 $$

$$ asin(t)cos(\theta)=bcos(t)sin(\theta) $$ $$ tan(t)=\frac{b}{a}tan(\theta) $$

Following the same steps for $y$ we find the $t$ that gives the maximum $y$:

$$ y=-acos(t)sin(\theta) + bsin(t)cos(\theta) $$

$$ \frac{dy}{dt}=asin(t)sin(\theta)+bcos(t)cos(\theta)=0 $$

$$ asin(t)sin(\theta)=-bcos(t)cos(\theta) $$

$$ tan(t)=-\frac{b}{a}cot(\theta) $$

ellipse boundary after rotation

2 Answers2