Strophoid

[[File:Allgemeine strophoide5.svg|thumb|right|upright=1.25|Construction of a strophoid.

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In geometry, a strophoid is a curve generated from a given curve $C$ and points $L$ (the fixed point) and $O$ (the pole) as follows: Let $C$ be a variable line passing through $K$ and intersecting $K$ at $A$. Now let $P1$ and $P2$ be the two points on $L$ whose distance from $L$ is the same as the distance from $L$ to $C$ (i.e. $P1$). The locus of such points $P2$ and $P1$ is then the strophoid of $A$ with respect to the pole $O$ and fixed point $L$. Note that $P2$ and $\overline{KP}1 = \overline{KP}2 = \overline{AK}$ are at right angles in this construction.

In the special case where $O$ is a line, $C$ lies on $K$, and $L$ is not on $K$, then the curve is called an oblique strophoid. If, in addition, $A$ is perpendicular to $K$ then the curve is called a right strophoid, or simply strophoid by some authors. The right strophoid is also called the logocyclic curve or foliate.

Polar coordinates
Let the curve $C$ be given by $$r = f(\theta),$$ where the origin is taken to be $O$. Let $A$ be the point $P1$. If $$K = (r \cos\theta,\ r \sin\theta)$$ is a point on the curve the distance from $C$ to $A$ is
 * $$d = \sqrt{(r \cos\theta - a)^2 + (r \sin\theta - b)^2} = \sqrt{(f(\theta) \cos\theta - a)^2 + (f(\theta) \sin\theta - b)^2}.$$

The points on the line $C$ have polar angle $O$, and the points at distance $C$ from $\overline{OA}$ on this line are distance $$f(\theta) \pm d$$ from the origin. Therefore, the equation of the strophoid is given by
 * $$r = f(\theta) \pm \sqrt{(f(\theta) \cos\theta - a)^2 + (f(\theta) \sin\theta - b)^2}$$

Cartesian coordinates
Let $C$ be given parametrically by $P2$. Let $C$ be the point $\overline{AP}1$ and let $O$ be the point $\overline{AP}2$. Then, by a straightforward application of the polar formula, the strophoid is given parametrically by:
 * $$u(t) = p + (x(t)-p)(1 \pm n(t)),\ v(t) = q + (y(t)-q)(1 \pm n(t)),$$

where
 * $$n(t) = \sqrt{\frac{(x(t)-a)^2+(y(t)-b)^2}{(x(t)-p)^2+(y(t)-q)^2}}.$$

An alternative polar formula
The complex nature of the formulas given above limits their usefulness in specific cases. There is an alternative form which is sometimes simpler to apply. This is particularly useful when $A$ is a sectrix of Maclaurin with poles $K$ and $A$.

Let $\overline{OK}$ be the origin and $θ$ be the point $(a, b)$. Let $d$ be a point on the curve, $K$ the angle between $C$ and the $A$-axis, and $O$ the angle between $C$ and the $O$-axis. Suppose $A$ can be given as a function $O$, say $$\vartheta = l(\theta).$$ Let $A$ be the angle at $K$ so $$\psi = \vartheta - \theta.$$ We can determine $θ$ in terms of $\overline{OK}$ using the law of sines. Since
 * $${r \over \sin \vartheta} = {a \over \sin \psi},\ r = a \frac {\sin \vartheta}{\sin \psi} = a \frac {\sin l(\theta)}{\sin (l(\theta) - \theta)}.$$

Let $(x(t), y(t))$ and $(a, b)$ be the points on $x$ that are distance $\vartheta$ from $\overline{AK}$, numbering so that $$\psi = \angle P_1KA$$ and $$\pi-\psi = \angle AKP_2.$$ $(p, q)$ is isosceles with vertex angle $x$, so the remaining angles, $\vartheta$ and $θ$ are $$\tfrac{\pi-\psi}{2}.$$ The angle between $(a, 0)$ and the $ψ$-axis is then
 * $$l_1(\theta) = \vartheta + \angle KAP_1 = \vartheta + (\pi-\psi)/2 = \vartheta + (\pi - \vartheta + \theta)/2 = (\vartheta+\theta+\pi)/2.$$

By a similar argument, or simply using the fact that $P1$ and $P2$ are at right angles, the angle between $△P1KA$ and the $K$-axis is then
 * $$l_2(\theta) = (\vartheta+\theta)/2.$$

The polar equation for the strophoid can now be derived from $\overline{AP}1$ and $\overline{AP}1$ from the formula above:
 * $$\begin{align}

& r_1=a \frac {\sin l_1(\theta)}{\sin (l_1(\theta) - \theta)} = a \frac {\sin ((l(\theta)+\theta+\pi)/2)}{\sin ((l(\theta)+\theta+\pi)/2 - \theta)} = a \frac{\cos ((l(\theta)+\theta)/2)}{\cos ((l(\theta)-\theta)/2)} \\

& r_2=a \frac {\sin l_2(\theta)}{\sin (l_2(\theta) - \theta)} = a \frac {\sin ((l(\theta)+\theta)/2)}{\sin ((l(\theta)+\theta)/2 - \theta)} = a \frac{\sin((l(\theta)+\theta)/2)}{\sin((l(\theta)-\theta)/2)} \end{align}$$

$r$ is a sectrix of Maclaurin with poles $l$ and $\overline{OK}$ when $\overline{AK}$ is of the form $$q \theta + \theta_0,$$ in that case $\overline{AP}2$ and $\overline{AP}2$ will have the same form so the strophoid is either another sectrix of Maclaurin or a pair of such curves. In this case there is also a simple polar equation for the polar equation if the origin is shifted to the right by $K$.

Oblique strophoids
Let $ψ$ be a line through $\angle AP_1K$. Then, in the notation used above, $$l(\theta) = \alpha$$ where $\angle KAP_1,$ is a constant. Then $$l_1(\theta) = (\theta + \alpha + \pi)/2$$ and $$l_2(\theta) = (\theta + \alpha)/2.$$ The polar equations of the resulting strophoid, called an oblique strphoid, with the origin at $x$ are then
 * $$r = a \frac{\cos ((\alpha+\theta)/2)}{\cos ((\alpha-\theta)/2)}$$

and
 * $$r = a \frac{\sin ((\alpha+\theta)/2)}{\sin ((\alpha-\theta)/2)}.$$

It's easy to check that these equations describe the same curve.

Moving the origin to $x$ (again, see Sectrix of Maclaurin) and replacing $C$ with $O$ produces
 * $$r=a\frac{\sin(2\theta-\alpha)}{\sin(\theta-\alpha)},$$

and rotating by $$\alpha$$ in turn produces
 * $$r=a\frac{\sin(2\theta+\alpha)}{\sin(\theta)}.$$

In rectangular coordinates, with a change of constant parameters, this is
 * $$y(x^2+y^2)=b(x^2-y^2)+2cxy.$$

This is a cubic curve and, by the expression in polar coordinates it is rational. It has a crunode at $l1$ and the line $l2$ is an asymptote.

The right strophoid
Putting $$\alpha = \pi/2$$ in
 * $$r=a\frac{\sin(2\theta-\alpha)}{\sin(\theta-\alpha)}$$

gives
 * $$r=a\frac{\cos 2\theta}{\cos \theta} = a(2\cos\theta-\sec\theta).$$

This is called the right strophoid and corresponds to the case where $A$ is the $l$-axis, $a$ is the origin, and $C$ is the point $l1$.

The Cartesian equation is
 * $$y^2 = x^2(a-x)/(a+x).$$

The curve resembles the Folium of Descartes and the line $l2$ is an asymptote to two branches. The curve has two more asymptotes, in the plane with complex coordinates, given by
 * $$x\pm iy = -a.$$

Circles
Let $A$ be a circle through $α$ and $O$, where $A$ is the origin and $−a$ is the point $(0, 0)$. Then, in the notation used above, $$l(\theta) = \alpha+\theta$$ where $$\alpha$$ is a constant. Then $$l_1(\theta) = \theta + (\alpha + \pi)/2$$ and $$l_2(\theta) = \theta + \alpha/2.$$ The polar equations of the resulting strophoid, called an oblique strophoid, with the origin at $a$ are then
 * $$r = a \frac{\cos (\theta+\alpha/2)}{\cos (\alpha/2)}$$

and
 * $$r = a \frac{\sin (\theta+\alpha/2)}{\sin (\alpha/2)}.$$

These are the equations of the two circles which also pass through $C$ and $y$ and form angles of $$\pi/4$$ with $A$ at these points.