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In geometry, a strophoid is a curve generated from a given curve C and points A (the fixed point) and O (the pole) as follows: Let L be a variable line passing through O and intersecting C at K. Now let P₁ and P₂ be the two points on L whose distance from K is the same as the distance from A to K. The locus of such points P₁ and P₂ is then the strophoid of C with respect to the pole O and fixed point A. Note that AP₁ and AP₂ are at right angles in this construction.

In the special case where C is a line, A lies on C, and O is not on C, then the curve is called an oblique strophoid. If, in addition, OA is perpendicular to C 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.

Equations

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 (a, b). If $K=(r\cos \theta ,\ r\sin \theta )$ is a point on the curve the distance from K 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 OK have polar angle $\theta$, and the points at distance d from K 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 (x(t), y(t)). Let A be the point (a, b) and let O be the point (p, q). 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 C is a sectrix of Maclaurin with poles O and A.

Let O be the origin and A be the point (a, 0). Let K be a point on the curve, $\theta$ the angle between OK and the x-axis, and $\vartheta$ the angle between AK and the x-axis. Suppose $\vartheta$ can be given as a function $\theta$, say $\vartheta =l(\theta )$. Let $\psi$ be the angle at K so $\psi =\vartheta -\theta$. We can determine r in terms of l 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 P₁ and P₂ be the points on OK that are distance AK from K, numbering so that $\psi =\angle P_{1}KA$ and $\pi -\psi =\angle AKP_{2}$. $\triangle P_{1}KA$ is isosceles with vertex angle $\psi$, so the remaining angles, $\angle AP_{1}K$ and $\angle KAP_{1}$, are $(\pi -\psi )/2$. The angle between AP₁ and the x-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 AP₁ and AP₂ are at right angles, the angle between AP₂ and the x-axis is then

$l_{2}(\theta )=(\vartheta +\theta )/2$.

The polar equation for the strophoid can now be derived from l₁ and l₂ from the formula above:

$r_{1}=a{\frac {\sin l_{1}(\theta )}{\sin(l_{1}(\theta )-\theta )}}=a{\frac {\sin((l(\theta )+\theta +\pi )/2)}{\sin((l(\theta )+\theta +\pi )/2-\theta )}}=a{\frac {\cos((l(\theta )+\theta )/2)}{\cos((l(\theta )-\theta )/2)}}$

$r_{2}=a{\frac {\sin l_{2}(\theta )}{\sin(l_{2}(\theta )-\theta )}}=a{\frac {\sin((l(\theta )+\theta )/2)}{\sin((l(\theta )+\theta )/2-\theta )}}=a{\frac {\sin((l(\theta )+\theta )/2)}{\sin((l(\theta )-\theta )/2)}}$

C is a sectrix of Maclaurin with poles O and A when l is of the form $q\theta +\theta _{0}$, in that case l₁ and l₂ 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 a.

Specific cases

Oblique strophoids

Let C be a line through A. Then, in the notation used above, $l(\theta )=\alpha$ 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 strphoid, with the origin at O 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 A (again, see Sectrix of Maclaurin) and replacing −a with a 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 (0, 0) and the line y=b 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 C is the y-axis, O is the origin, and A is the point (a,0).

The Cartesian equation is

$y^{2}=x^{2}(a-x)/(a+x)$.

The curve resembles the Folium of Descartes and the line x = −a 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 C be a circle through O and A, where O is the origin and A is the point (a, 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 strphoid, with the origin at O 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 O and A and form angles of $\pi /4$ with C at these points.

References

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• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
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• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
• "Courbe Strophoïdale" at Encyclopédie des Formes Mathématiques Remarquables (in French)
• "Strophoïde" at Encyclopédie des Formes Mathématiques Remarquables (in French)
• "Strophoïde Droite" at Encyclopédie des Formes Mathématiques Remarquables (in French)
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