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 $(a,b)$.
On
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 $y$:
$$ y=-acos(t)sin(\theta) + bsin(t)cos(\theta) $$
$$ \frac{dy}{dt}=asin(t)sin(\theta)+bcos(t)cos(\theta)=0 $$
$$ asin(t)sin(\theta)=-bcos(t)cos(\theta) $$
$$ tan(t)=-\frac{b}{a}cot(\theta) $$
$$ r(t)=(a\,\cos (t), b\, \sin(t)) $$
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?