Orthoptic (geometry)

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In the geometry of curves, an orthoptic is the set of points for which two tangents of a given curve meet at a right angle.

The orthoptic of a parabola is its directrix.
Ellipse and its orthoptic (purple)
Hyperbola with its orthoptic (purple)
Examples

The orthoptic of

1) a parabola is its directrix (proof: s. parabola),
2) an ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is the director circle x^2+y^2=a^2+b^2 (s. below),
3) a hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \, a>b, is the circle x^2+y^2=a^2-b^2 (in case of a\le b there are no orthogonal tangents, s. below),
4) an astroid x^{2/3} + y^{2/3}=1 is a quadrifolium with the polar equation
r=\tfrac{1}{\sqrt{2}}\cos(2\varphi), \ 0\le \varphi <2\pi, (s. below).
Generalizations
1) An isoptic is the set of points for which two tangents of a given curve meet at a fixed angle (s. below).
2) An isoptic of two plane curves is the set of points for which two tangents meet at a fixed angle.
3) Thales' theorem on a chord \overline{P_1P_2} can be considered as the orthoptic of two circles which are degenerated to the two points P_1,P_2.

Remark: In medicine there exists the term orthoptic, too.

Orthoptic of an ellipse and hyperbola[edit]

Ellipse[edit]

Main article: Director circle

The ellipse with equation \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 can be represented by the unusual parametric representation[1]

  • \vec c_\pm(m)=\left(-\frac{ma^2}{\pm\sqrt{m^2a^2+b^2}},\frac{b^2}{\pm\sqrt{m^2a^2+b^2}}\right),\, m \in \R,

where m is the slope of the tangent at a point of the ellipse. \vec c_+(m) describes the upper half and \vec c_-(m) the lower half of the ellipse. The points (\pm a,0)) with tangents parallel to the y-axis are excluded. But this is no problem, because these tangents meet orthogonal the tangents parallel to the x-axis in the ellipse points (0, \pm b) Hence the points (\pm a, \pm b) are points of the desired orthoptic (circle x^2+y^2=a^2+b^2).

The tangent at point \vec c_\pm(m) has the equation

y=m x  \pm\sqrt{m^2a^2+b^2}.

If a tangent contains the point (x_0,y_0), off the ellipse, then the equation

y_0=m x_0  \pm \sqrt{m^2a^2+b^2}

holds. Eliminatig the square root leads to

m^2-\frac{2x_0y_0}{x_0^2-a^2}m + \frac{y_0^2-b^2}{x_0^2-a^2}=0,

which has two solutions m_1,m_2 corresponding to the two tangents passing point (x_0,y_0). The free term of a reduced quadratic equation is always the product of its solutions. Hence, if the tangents meet at point (x_0,y_0) orthogonal, the follwing equations hold:

m_1m_2=-1=\frac{y_0^2-b^2}{x_0^2-a^2}

The last equation is equivalent to

x_0^2+y_0^2=a^2+b^2

This means:

  • The intersection points of orthogonal tangents are points of the circle x^2+y^2=a^2+b^2.

Hyperbola[edit]

The ellipse case can be adopted nearly literally to the hyperbola case. The only changes to be made are: 1) replace b^2 by -b^2 and 2) restrict m by |m|>b/a. One gets:

  • The intersection points of orthogonal tangents are points of the circle x^2+y^2=a^2-b^2,\ a>b .

Orthoptic of an astroid[edit]

Orthoptic (purple) of an astroid

An astroid can be described by the parametric representation

\vec c(t)=(\cos^3t,\sin^3t), \; 0\le t<2\pi.

From the condition \vec \dot c(t)\cdot\vec \dot c(t+\alpha)=0 one recognizes, at what distance \alpha in parameter space an orthogonal tangent to \vec \dot c(t) appears. It turns out, that the distance is independent of parameter t, namely \alpha=\pm \tfrac{\pi}{2}. The equations of the (orthogonal) tangents at the points \vec c(t) and \vec c(t+\tfrac{\pi}{2}) are:

y=-\tan t (x-\cos^3 t)+\sin^3t, \ y=\tfrac{1}{\tan t} (x+\sin^3 t)+\cos^3t.

There common point has coordinates:

x=\sin t\cos t(\sin t-\cos t),
y=\sin t\cos t(\sin t+\cos t).

This is at the same time a parametric representation of the orthoptic.

Elimination of parameter t yields the implicit representation

2(x^2+y^2)^3-(x^2-y^2)^2=0.

Introducing the new parameter \varphi=t-\tfrac{5}{4}\pi one gets

x=\tfrac{1}{\sqrt{2}}\cos(2\varphi)\,\cos\varphi, \ \ y=\tfrac{1}{\sqrt{2}}\cos(2\varphi)\,\sin\varphi. (proof uses Angle sum and difference identities.)

Herefrom we get the polar representation

r=\frac{1}{\sqrt{2}}\cos(2\varphi), \ 0\le \varphi <2\pi

of the orthoptic. Hence

Isoptic of a parabola, an ellipse and a hyperbola[edit]

Isoptics (purple) of a parabola for angles 80 and 100 degree
Isoptics (purple) of an ellipse for angles 80 and 100 degree
Isoptics (purple) of a hyperbola for angles 80 and 100 degree

Below isotopics for an angle \alpha\ne 90^\circ are listed. They are called \alpha-isoptics. For the proofs: s. below.

Equations of the isoptics[edit]

parabola :

The \alpha-isoptics of the parabola with equation y=ax^2 are the arms of the hyperbola

x^2-\tan^2\alpha\left(y+\frac{1}{4a}\right)^2-\frac{y}{a}=0.

The arms of the hyperbola provide the isoptics for the two angles \alpha, 180^\circ\!-\!\alpha (s. picture).

Ellipse:

The \alpha-isoptics of the ellipse with equation \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 are the two parts of the curve of 4. degree

\tan^2\alpha\;(x^2+y^2-a^2-b^2)^2=4(a^2y^2+b^2x^2-a^2b^2) (s. picture).
Hyperbola:

The \alpha-isoptics of the hyperbola with the equation \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 are the two parts of the curve of 4. degree

\tan^2\alpha\;(x^2+y^2-a^2+b^2)^2=4(a^2y^2-b^2x^2+a^2b^2).

Proofs[edit]

Parabola:

A parabola y=ax^2 can be parametrized by the slope of its tangents m=2ax:

\vec c(m)=\left(\frac{m}{2a},\frac{m^2}{4a}\right), \, m \in \R.

The tangent with slope m has the equation

y=mx-\frac{m^2}{4a}.

Point (x_0,y_0) is on the tangent, if

y_0=mx_0-\frac{m^2}{4a}

holds. That means, the slopes m_1,m_2 of the two tangents containing (x_0,y_0) fulfill the quadratic equation

m^2 - 4ax_0m + 4ay_0=0.

If the tangents meet with angle \alpha or 180^\circ -\alpha, the equation

\tan^2\alpha=\left(\frac{m_1-m_2}{1+m_1m_2}\right)^2

has to be fulfilled. Solving the quadratic equation for m, inserting m_1,m_2 into the last equation, one gets

x_0^2-\tan^2\alpha\left(y_0+\frac{1}{4a}\right)^2-\frac{y_0}{a}=0.

This is the equation of the hyperbola above. Its arms bear the two isoptics of the parabola for the two angles \alpha and 180^\circ-\alpha.

Ellipse:

In case of an ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 one can adopt the idea for the orthoptic until the quadratic equation

m^2-\frac{2x_0y_0}{x_0^2-a^2}m + \frac{y_0^2-b^2}{x_0^2-a^2}=0.

Now, as in case of a parabola, the quadratic equation has to be solved and the two solutions m_1,m_2 to be inserted into the equation \tan^2\alpha=\left(\tfrac{m_1-m_2}{1+m_1m_2}\right)^2. Rearranging shows that the isoptics are parts of the curve of 4. degree:

\tan^2\alpha\;(x_0^2+y_0^2-a^2-b^2)^2=4(a^2y_0^2+b^2x_0^2-a^2b^2).
Hyperbola:

The solution for the case of a hyperbola can be adopted from the ellipse case by replacing b^2 by -b^2 (as in the case of the orthoptics, s. above).

Remark: For visualizing the isoptics see implicit curve.

External links[edit]

Notes[edit]

  1. ^ P. K. Jain: Textbook of Analytical Geometry of two Dimensions. p. 214.

References[edit]

  • J. Dennis Lawrence (1972). A catalog of special plane curves. Dover Publications. pp. 58–59. ISBN 0-486-60288-5. 
  • Boris Odehnal: Equioptic Curves of Conic Sections. Journal for Geometry and Graphics, Volume 14 (2010), No. 1, p. 29–43.
  • Hermann Schaal: Lineare Algebra und Analytische Geometrie. Band III, Vieweg, 1977, ISBN 3 528 03058 5, p. 220.
  • Jacob Steiner’s Vorlesungen über synthetische Geometrie. B. G. Teubner, Leipzig 1867 (bei Google Books), 2. Teil, p. 186.
  • Maurizio Ternullo: Two new sets of ellipse related concyclic points. Journal of Geometry 2009, 94, p. 159–173.