Rytz's construction

(1) rotate point ${\displaystyle P}$ around ${\displaystyle C}$ by 90°.
(2) Determine the center ${\displaystyle D}$ of the line segment ${\displaystyle {\overline {P'Q}}}$.
(3) Draw the line ${\displaystyle P'Q}$ and the circle with center ${\displaystyle D}$ through ${\displaystyle C}$. Intersect the circle and the line. The intersection points are ${\displaystyle A,B}$.
(4) The lines ${\displaystyle CA}$ and ${\displaystyle CB}$ are the axes of the ellipse.
(5) The line segment ${\displaystyle {\overline {AB}}}$ can be considered as a paperstrip of length ${\displaystyle a+b}$ (see ellipse) generating point ${\displaystyle Q}$. Hence ${\displaystyle a=|AQ|}$ and ${\displaystyle b=|BQ|}$ are the semi-axes. (If ${\displaystyle a>b}$ then ${\displaystyle a}$ is the semi-major axis.)
(6) The vertices and co-vertices are known and the ellipse can be drawn by one of the drawing methods.

If one performs a left turn of point ${\displaystyle P}$, then the configuration shows the 2nd paper strip method (see second diagram in next section) and ${\displaystyle a=|AQ|}$ and ${\displaystyle b=|BQ|}$ is still true.

(1): Any ellipse can be represented in a suitable coordinate system parametrically by

${\displaystyle {\vec {p}}(t)=(a\cos t,\;b\sin t)^{T}}$.

Two points ${\displaystyle {\vec {p}}(t_{1}),\ {\vec {p}}(t_{2})}$ lie on conjugate diameters if ${\displaystyle t_{2}-t_{1}=\pm {\tfrac {\pi }{2}}\ .}$ (see Ellipse: conjugate diameters.)

(2): Let be ${\displaystyle Q=(a\cos t,\;b\sin t)}$ and

${\displaystyle P=(a\cos(t+{\tfrac {\pi }{2}}),\;b\sin(t+{\tfrac {\pi }{2}}))=(-a\sin t,b\cos t)}$

two points on conjugate diameters.

Then ${\displaystyle P'=(b\cos t,a\sin t)}$ and the midpoint of line segment ${\displaystyle {\overline {P'Q}}}$ is ${\displaystyle D=\left({\tfrac {a+b}{2}}\cos t,{\tfrac {a+b}{2}}\sin t\right)}$.

(3): Line ${\displaystyle P'Q}$ has equation ${\displaystyle x\sin t+y\cos t=(a+b)\;\cos t\sin t\ .}$

The intersection points of this line with the axes of the ellipse are

${\displaystyle U=\left(0,\;(a+b)\sin t\right)\ ,\quad V=\left((a+b)\cos t,\;0\right)\ .}$

Rytz: left turn of point ${\displaystyle P}$

(4): Because of ${\displaystyle |UD|=|VD|=|CD|}$ the points ${\displaystyle U,V,C}$ lie on the circle with center ${\displaystyle D}$ and radius ${\displaystyle |CD|\ .}$

Hence ${\displaystyle A=U,\ B=V\ .}$

(5): ${\displaystyle |UQ|=a,\quad |VQ|=b\ .}$

The proof uses a right turn of point ${\displaystyle P}$, which leads to a diagram showing the 1st paper strip method.

If one performs a left turn of point ${\displaystyle P}$, then results (4) and (5) are still valid and the configuration shows now the 2nd paper strip method (see diagram).
If one uses ${\displaystyle P=(a\cos(t{\color {red}-}{\tfrac {\pi }{2}}),\;b\sin(t{\color {red}-}{\tfrac {\pi }{2}}))=(a\sin t,-b\cos t)}$, then the construction and proof work either.