# ellipse boundary after rotation

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

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:

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 $E_\theta= R_\theta (E)$? Commented Oct 12, 2013 at 9:15
• 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). Commented Oct 12, 2013 at 9:21

$$r(t)=(a\,\cos (t), b\, \sin(t))$$
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?

• I don't understand what $t$ is? Commented Oct 12, 2013 at 9:31
• $t\in[0,2\pi]$ and $r(t)$ is a parametrization of the ellipse. Commented Oct 12, 2013 at 9:33
• Do you have some knowledge of calculus? Commented Oct 12, 2013 at 9:35
• 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. Commented Oct 12, 2013 at 9:45
• You are welcome. There is an exact analytic solution but is bit complicated. Commented Oct 12, 2013 at 10:07

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 $y$:

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

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

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

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