Strophoid

Polar coordinates Let the curve be given by r = f(\theta), where the origin is taken to be . Let be the point . If K = (r \cos\theta,\ r \sin\theta) is a point on the curve the distance from to 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 have polar angle , and the points at distance from 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 be given parametrically by . Let be the point and let be the point . 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 is a sectrix of Maclaurin with poles and . Let be the origin and be the point . Let be a point on the curve, the angle between and the -axis, and the angle between and the -axis. Suppose can be given as a function , say \vartheta = l(\theta). Let be the angle at so \psi = \vartheta - \theta. We can determine in terms of 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 and be the points on that are distance from , numbering so that \psi = \angle P_1KA and \pi-\psi = \angle AKP_2. is isosceles with vertex angle , so the remaining angles, and are \tfrac{\pi-\psi}{2}. The angle between and the -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 and are at right angles, the angle between and the -axis is then :l_2(\theta) = (\vartheta+\theta)/2. The polar equation for the strophoid can now be derived from and 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} is a sectrix of Maclaurin with poles and when is of the form q \theta + \theta_0, in that case and 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 . ==Specific cases==

Strophoids of lines are actually expressible as singular cubics in the projective plane. Oblique strophoids Let be a line through . Then, in the notation used above, l(\theta) = \alpha where 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 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 (again, see Sectrix of Maclaurin) and replacing with 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 and the line 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 is the -axis, is the origin, and is the point . The Cartesian equation is :y^2 = x^2(a-x)/(a+x). The curve resembles the Folium of Descartes and the line is an asymptote to two branches. The curve has two more asymptotes, in the plane with complex coordinates, given by :x\pm iy = -a. This curve passes through the two circular points at infinity and is a special case of a focal circular Van Rees cubic. Circles Let be a circle through and , where is the origin and is the point . 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 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 and and form angles of \pi/4 with at these points. ==See also==