<s>[INST] How to calculate width and height of a 45° rotated ellipse bounded by a square? I'm coming from a programming background so I apologies if this is blindingly simple or I misuse terms. I have an ellipse bounded by a square. For simplicity the centre of the square and ellipse is the origin (0,0) while the square is 2 width and 2 height.
The ellipse is rotated -45° or +45° (angle in image) and I can easily work this out. The ellipse touches all sides of the square, and I also know the intersection points. In the image A is the distance between the corner and the intersection while B is the length of the long section (the other points are symmetric).

What is the width and height, as described in the image, of the ellipse?
 [/INST] Consider an square formed by the sides of $\pm x\pm y=m$ (i.e. side length $m\sqrt2$). 
Now find an ellipse which touches the square. Consider the top right quadrant. Edge of square is $$L: \quad x+y=m$$Assume point of tangency is $P(h,k)$. 
Ellipse, $E$:
$$\begin{align}
\frac {x^2}{a^2}+\frac {y^2}{b^2}&=1\\
\frac {dy}{dx}&=-\frac {b^2x}{a^2y}=-\frac {b^2h}{a^2k}=-1\quad\text{at }P\\
\frac h{a^2}&=\frac k{b^2}
\end{align}$$
$P$ lies on L, hence $h+k=m$, which gives
$$\begin{align}
h&=\left(\frac {m}{a^2+b^2}\right)a^2\\
k&=\left(\frac {m}{a^2+b^2}\right)b^2
\end{align}$$
$P$ lies on $E$, hence
$$\begin{align}
\left(\frac {m}{a^2+b^2}\right)^2\left[\frac {(a^2)^2}{a^2}+\frac {(b^2)^2}{b^2}\right]&=1\\
m&=\sqrt{a^2+b^2}
\end{align}$$
This gives 
$$\begin{align}
h&=\frac {a^2}m\\
k&=\frac {b^2}m\end{align}$$
$P$ divides edge of square into side lengths $$B=h\sqrt{2}=\frac {a^2\sqrt{2}}m\\ A=k\sqrt{2}=\frac {b^2\sqrt{2}}m$$ 
In the question posted, side length of square is $2$, hence $m=\sqrt{2}$, which gives
$$A=b^2\\
B=a^2$$
"Height" (major axis) and "Width" (minor axis) of $E$ are  given by
$$\text{Height (major axis)=}\color{red}{2a=2\sqrt{B}}\\
\text{Width (minor axis)=}\color{red}{2b=2\sqrt{A}}$$

Note
$\hspace{3cm}$ 
From this solution to another recent MSE question, note that the ellipse 
$$\frac{x^2}{u}+\frac {y^2}{1-u}=1$$
has a nice property in that its point of tangency with the  line $L: x+y=1$ is  $(u,1-u)$ which divides the line segment $V(0,1), H(1,0)$ on $L$ in the ratio $u:1-u$. Note that $0<u<1$. The line segment $VH$ has length $\sqrt2$. The semi-major and semi-minor axes of the ellipse are $\sqrt{u}, \sqrt{1-u}$ respectively. 
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