Strophoid

Construction of a strophoid. Template:Legend-line Template:Legend-line Template:Legend-line Template:Legend-line Template:Legend-line

In geometry, a strophoid is a curve generated from a given curve Template:Mvar and points Template:Mvar (the fixed point) and Template:Mvar (the pole) as follows: Let Template:Mvar be a variable line passing through Template:Mvar and intersecting Template:Mvar at Template:Mvar. Now let Template:Math and Template:Math be the two points on Template:Mvar whose distance from Template:Mvar is the same as the distance from Template:Mvar to Template:Mvar (i.e. Template:Math). The locus of such points Template:Math and Template:Math is then the strophoid of Template:Mvar with respect to the pole Template:Mvar and fixed point Template:Mvar. Note that Template:Math and Template:Math are at right angles in this construction.

In the special case where Template:Mvar is a line, Template:Mvar lies on Template:Mvar, and Template:Mvar is not on Template:Mvar, then the curve is called an oblique strophoid. If, in addition, Template:Mvar is perpendicular to Template:Mvar 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 Template:Mvar be given by $r=f(\theta ),$ where the origin is taken to be Template:Mvar. Let Template:Mvar be the point Template:Math. If $K=(r\cos \theta ,r\sin \theta )$ is a point on the curve the distance from Template:Mvar to Template:Mvar 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 Template:Mvar have polar angle Template:Mvar, and the points at distance Template:Mvar from Template:Mvar 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 Template:Mvar be given parametrically by Template:Math. Let Template:Mvar be the point Template:Math and let Template:Mvar be the point Template:Math. 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 Template:Mvar is a sectrix of Maclaurin with poles Template:Mvar and Template:Mvar.

Let Template:Mvar be the origin and Template:Mvar be the point Template:Math. Let Template:Mvar be a point on the curve, Template:Mvar the angle between Template:Mvar and the Template:Mvar-axis, and Template:Tmath the angle between Template:Mvar and the Template:Mvar-axis. Suppose Template:Tmath can be given as a function Template:Mvar, say $\vartheta =l(\theta ).$ Let Template:Mvar be the angle at Template:Mvar so $\psi =\vartheta -\theta .$ We can determine Template:Mvar in terms of Template:Mvar using the law of sines. Since

$\frac{r}{\sin \vartheta }=\frac{a}{\sin \psi },r=a\frac{\sin \vartheta }{\sin \psi }=a\frac{\sin l(\theta )}{\sin (l(\theta )-\theta )}.$

Let Template:Math and Template:Math be the points on Template:Mvar that are distance Template:Mvar from Template:Mvar, numbering so that $\psi =\mathrm{\angle }{P}_{1}KA$ and $\pi -\psi =\mathrm{\angle }AK{P}_2.$ Template:Math is isosceles with vertex angle Template:Mvar, so the remaining angles, Template:Tmath and Template:Tmath are $\frac{\pi -\psi }{2}.$ The angle between Template:Math and the Template:Mvar-axis is then

$l_1(\theta )=\vartheta +\mathrm{\angle }KA{P}_1=\vartheta +(\pi -\psi )/2=\vartheta +(\pi -\vartheta +\theta )/2=(\vartheta +\theta +\pi )/2.$

By a similar argument, or simply using the fact that Template:Math and Template:Math are at right angles, the angle between Template:Math and the Template:Mvar-axis is then

$l_2(\theta )=(\vartheta +\theta )/2.$

The polar equation for the strophoid can now be derived from Template:Math and Template:Math from the formula above:

$\begin{array}{cc}& r_1=a\frac{\sin l_1(\theta )}{\sin (l_1(\theta )-\theta )}=a\frac{\sin ((l(\theta )+\theta +\pi )/2)}{\sin ((l(\theta )+\theta +\pi )/2-\theta )}=a\frac{\cos ((l(\theta )+\theta )/2)}{\cos ((l(\theta )-\theta )/2)}\\ & r_2=a\frac{\sin l_2(\theta )}{\sin (l_2(\theta )-\theta )}=a\frac{\sin ((l(\theta )+\theta )/2)}{\sin ((l(\theta )+\theta )/2-\theta )}=a\frac{\sin ((l(\theta )+\theta )/2)}{\sin ((l(\theta )-\theta )/2)}\end{array}$

Template:Mvar is a sectrix of Maclaurin with poles Template:Mvar and Template:Mvar when Template:Mvar is of the form $q\theta +{\theta }_0,$ in that case Template:Math and Template:Math 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 Template:Mvar.

Specific cases

Oblique strophoids

Let Template:Mvar be a line through Template:Mvar. Then, in the notation used above, $l(\theta )=\alpha$ where Template:Mvar 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 Template:Mvar 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 Template:Mvar (again, see Sectrix of Maclaurin) and replacing Template:Mvar with Template:Mvar 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 Template:Math and the line Template:Math is an asymptote.

The right strophoid

A 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 Template:Mvar is the Template:Mvar-axis, Template:Mvar is the origin, and Template:Mvar is the point Template:Math.

The Cartesian equation is

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

The curve resembles the Folium of Descartes and the line Template:Math 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 Template:Mvar be a circle through Template:Mvar and Template:Mvar, where Template:Mvar is the origin and Template:Mvar is the point Template:Math. 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 Template:Mvar 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 Template:Mvar and Template:Mvar and form angles of $\pi /4$ with Template:Mvar at these points.

See also

• Conchoid
• Cissoid

References

• Template:Cite book
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• Template:MathWorld
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• Template:Springer
• Template:MacTutor