It is possible to construct an ellipse or a hyperbola by tracing the intersections of offset lines that rotate at the same rate.
Firstly, here's a graph that shows the rotating lines, as well as the tracing of their intersections: https://www.desmos.com/calculator/bjpuuy1i4m. Play the slider $t$ to see the tracing.
Two lines rotate in opposite directions around a single point. At another point, two other lines do the same, but with a 90° offset to the first two lines.
These lines are represented by these four equations:
$y=\tan(t)(x-u)$
$y=-\tan(t)(x-u)$
$y=\cot(t)(x+u)$
$y=-\cot(t)(x+u)$
Where $t$ is the angle of rotation and $u$ is the x-intercept of the line.
For any value of $t$, all intersections between every line is a either a point on the circle $x^2+y^2=u^2$ or a point on the hyperbola $x^2-y^2=u^2$, as long as $u≠0$ .
Why do all of the intersections always fall on a circle or a hyperbola, and how might this construction connect to some of the other constructions of conics?
Best Answer
Consider a line $r$ of slope $m$ passing through $A=(-u,0)$: its equation is $y=m(x+u)$.
Line $s$, perpendicular to $r$ and passing through $B=(u,0)$, has equation $y=(-1/m)(x-u)$. Lines $r$ and $s$ intersect at $P=\left({1-m^2\over1+m^2}u,{2m\over1+m^2}u\right)$ and by Thale's theorem $P$ lies on the circle of diameter $AB$ (you can also check that $x_P^2+y_P^2=u^2$).
Line $t$, reflection of $s$ about line $x=u$, has equation $y=(1/m)(x-u)$ and intersects $r$ at $Q=\left({1+m^2\over1-m^2}u,{2m\over1-m^2}u\right)$. You can check that $x_Q^2-y_Q^2=u^2$, hence $Q$ lies on the equilateral hyperbola having $A$ and $B$ as vertices.
As $m$ varies in $(-\infty,+\infty)$, points $P$ and $Q$ thus describe a circle and a hyperbola, as explained above. In the equations you report, the convenient parameterization $m=\tan t$ is used, but that is not necessary. In addition, a fourth line $r'$ is added: it does nothing new, but meets $s$ on the hyperbola and $t$ on the circle, as you can easily check.