Ellipse generated by a vertex of a sliding square on the $x$ and $y$ axes

Question

A unit square is placed with the bottom left corner at the origin and the top right corner at $(1,1)$. The base vertices then slide on the $x$ and $y$ axes as shown in the animation below. The top right vertex traces a curve. Show that this curve is an ellipse, and find its semi-major and semi-minor axes.

My attempt:

So, obviously, we want to parametrize the coordinates of the top right corner in terms of the rotation angle of the square. Turning the square by an angle $\theta$, the coordinates of the base are $(\cos \theta, 0)$ and $(0, \sin \theta)$, so the coordinates of the top right corner are consequently,

$ P = (\cos \theta, 0) + (\sin \theta, \cos \theta ) = ( \cos \theta + \sin \theta, \cos \theta) $

This can be written as

$P = \cos \theta (1, 1) + \sin \theta (1, 0) $

or more compactly as

$ P = \begin{bmatrix} 1 && 1 \\ 1 && 0 \end{bmatrix} u $

where $u = [\cos \theta, \sin \theta ]^T $

From this last equation, and solving for $u$, we get

$u = \begin{bmatrix} 1 && 1 \\ 1 && 0 \end{bmatrix}^{-1} P = \begin{bmatrix} 0 && 1 \\ 1 && -1 \end{bmatrix} P $

Now $u$ is a unit vector, i.e. $u^T u = 1$. Using the expression for $u$ in terms of $P$ we can write

$u^T u = 1 = P^T \begin{bmatrix} 0 && 1 \\ 1 && -1 \end{bmatrix}^T \begin{bmatrix} 0 && 1 \\ 1 && -1 \end{bmatrix} P = P^T \begin{bmatrix} 0 && 1 \\ 1 && -1 \end{bmatrix} \begin{bmatrix} 0 && 1 \\ 1 && -1 \end{bmatrix} P = P^T \begin{bmatrix} 1 && -1 \\ -1 && 2 \end{bmatrix} P $

Thus, we now have

$P^T Q P = 1 $

where $Q =\begin{bmatrix} 1 && -1 \\ -1 && 2 \end{bmatrix} $

Diagonalize $Q$ (i.e. put it in the form $Q = R D R^T$ as follows:

1. Define $\phi = \frac{1}{2} \tan^{-1}\left( \dfrac{2(-1)}{1 - 2} \right) = \frac{1}{2} \tan^{-1}(2) $

2. It follows that $ \cos(2 \phi) = \dfrac{1}{\sqrt{5}} \Rightarrow \cos(\phi) = \sqrt{ \dfrac{ 1 + \sqrt{5}}{2 \sqrt{5}}} \Rightarrow \sin(\phi) = \sqrt{\dfrac{\sqrt{5} - 1}{2 \sqrt{5}}} $

3. Step 2. above gives us the rotation matrix $R$, namely,

$ R = \begin{bmatrix} \cos(\phi) && - \sin(\phi) \\ \sin(\phi) && \cos(\phi) \end{bmatrix} $

3. Diagonal elements of $D$ are

$D_{11} = Q_{11} \cos^2(\phi) + Q_{22} \sin^2(\phi) + 2 Q_{12} \sin \phi \cos \phi = \dfrac{3 \sqrt{5} - 5}{2 \sqrt{5}} $

$D_{22} = Q_{11} \sin^2(\phi) + Q_{22} \cos^2(\phi) - 2 Q_{12} \sin \phi \cos \phi = \dfrac{5 + 3 \sqrt{5}}{2 \sqrt{5}} $

Hence, finally, we have the semi-major axis length

$a = \dfrac{1}{\sqrt{D_{11}}} = \sqrt{ \dfrac{ 2 \sqrt{5} }{3 \sqrt{5} - 5}} = \sqrt{ \dfrac{ 2 \sqrt{5} ( 3 \sqrt{5} + 5 ) }{20}}\\ = \sqrt{\dfrac{3 + \sqrt{5}}{2}} = \sqrt{ 1 + \Phi } = \Phi \approx 1.618$

and the semi-minor axis length

$b = \dfrac{1}{\sqrt{D_{22}}} = \sqrt{ \dfrac{ 2 \sqrt{5}}{ 5 + 3\sqrt{5}} } = \sqrt{ \dfrac{ 2 \sqrt{5}( 3 \sqrt{5} - 5 ) }{20} } = \\ \sqrt{ \dfrac{3 - \sqrt{5}}{2} } = \sqrt{ 2 - \Phi } = \sqrt{1 + 1 - \Phi} =\sqrt{1 + \Phi^2 - 2 \Phi } = \Phi - 1 \approx 0.618 $

Where $\Phi$ is the golden ratio, $\Phi = \dfrac{ 1 + \sqrt{5} }{2} $

Answer 1

$$x=\cos t+\sin t, y=\cos t$$ $$(x-y)^2+y^2=1$$ This is second order curve. $-2 \leq x \leq 2$, $-1 \leq y \leq 1$, then curve is limited, then it is ellipse.

Claim. Minimum and maximum value of $a\sin t + b\cos t$ are $\pm\sqrt{a^2+b^2}$. It is easy to prove.

$x_{max}=\sqrt{2}$, $x_{min}=-\sqrt{2}$, $y_{min}=-1$, $y_{max}=1$ then center is $$\left(\frac{x_{min}+x_{max}}{2};\frac{y_{min}+y_{max}}{2}\right)=(0;0)$$ Distance of ellipse point from center is $$\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}=\sqrt{1+2\sin t \cos t + \cos^2 t}=\sqrt{\frac{3}{2}+\sin 2t+\frac{1}{2}\cos 2t}$$ Maximum distance is $$a=\sqrt{\frac{3}{2}+\sqrt{1+\frac{1}{4}}}=\sqrt{\frac{3+\sqrt{5}}{2}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\frac{1+\sqrt{5}}{2}$$ minimum distance is $$b=\sqrt{\frac{3-\sqrt{5}}{2}}=\frac{\sqrt{5}-1}{2}$$