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)

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).
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.

Orthoptics (red circles) of a circle, ellipses and hyperbolas

Orthoptic of an ellipse and hyperbola[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 following equations hold:


The last equation is equivalent to


This means:

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


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


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


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


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).

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




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


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


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


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


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.


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:


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]


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


  • 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.