# Largest rotated ellipse inscribed in a rectangle

Let's say I have a parametrized ellipse

$$x (t) = a \cos(t) \cos(r) - b \sin(t) \sin(r)$$

$$y (t) = a \cos(t) \sin(r) + b \sin(t) \cos(r)$$

Where $r$ is the rotation around the axis and $t \in [0,2\pi]$ the parameter. How would I find the $a$ and $b$ such as the ellipse is largest, area wise, inside a rectangle of $w$ width and $h$ height?

Thanks math gurus!

Update: Both ellipse and rectangle are centered on $(0,0)$. Rotation $r \in [0,\pi/2]$ is elevation from the $x$ axis.

• Geometrically, $r$ is rotation around what axis? And given the axis, what does the angle $r$ mean with respect to that axis? Jun 22 '16 at 0:38
• We are only aloud to adjust $a$, and $b$, right? Does $r$ remain constant?
– Hrhm
Jun 22 '16 at 1:17
• I think that talking about the parametrization is a red herring, since this is a purely geometric problem. The question is on the correctness of the intuition that the maximizing ellipse is tangent to the rectangle at the midpoints of its sides. Jun 22 '16 at 1:53

The maxima of $x$ and $y$ when $t$ is free are

$$\hat x=\sqrt{(a\cos(r))^2+(b\sin(r))^2},\\\hat y=\sqrt{(a\sin(r))^2+(b\cos(r))^2}.$$

We need to express that they equal $w$ and $h$, and solve for $a$ and $b$, then maximize the product.

$$a^2+b^2=w^2+h^2,\\(a^2-b^2)\cos(2r)=w^2-h^2$$ and finally, subtracting the squares

$$4a^2b^2=(w^2+h^2)^2-\left(\frac{w^2-h^2}{\cos(2r)}\right)^2,$$ which is maximum for $r=0$, the axis-aligned ellipse.

• Those are the maxima along the axes of the ellipse, not necessarily along the $x$ and $y$ axis.
– Hrhm
Jun 22 '16 at 13:17
• @Hrhm: I think you are wrong. Please elaborate.
– user65203
Jun 22 '16 at 13:22
• Yep, I'm wrong, my bad
– Hrhm
Jun 22 '16 at 13:32
• I am sorry, I don't get it, you skipped too many steps for my little head. Could you please add a bit more details? e.g. how did you get the x and y maxima for starters ? Jun 22 '16 at 23:57
• @jldupont: use $p\cos(t)+q\sin(t)=\sqrt{p^2+q^2}\cos(t-\phi)$ to get the extrema. The rest is routine work.
– user65203
Jun 23 '16 at 6:28