Orthoptic (geometry)

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title: "Orthoptic (geometry)" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["curves"] description: "All points for which two tangents of a curve intersect at 90° angles" topic_path: "general/curves" source: "https://en.wikipedia.org/wiki/Orthoptic_(geometry)" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0

::summary All points for which two tangents of a curve intersect at 90° angles ::

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.

::figure[src="https://upload.wikimedia.org/wikipedia/commons/4/4e/Parabel-orthop.svg" caption=""] ::

]] ::figure[src="https://upload.wikimedia.org/wikipedia/commons/4/4b/Orthoptic-ellipse-s.svg" caption="[[Ellipse]]}}"] ::

]] ::figure[src="https://upload.wikimedia.org/wikipedia/commons/0/0a/Orthoptic-hyperbola-s.svg" caption="[[Hyperbola]]}}"] ::

]]

Examples:

1. The orthoptic of a parabola is its directrix (proof: see below),
2. The orthoptic of an ellipse \tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1 is the director circle x^2 + y^2 = a^2 + b^2 (see below),
3. The orthoptic of a hyperbola \tfrac{x^2}{a^2} - \tfrac{y^2}{b^2} = 1,\ a b is the director circle x^2 + y^2 = a^2 - b^2 (in case of a ≤ b there are no orthogonal tangents, see below),
4. The orthoptic of 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 (see below).

Generalizations:

1. An isoptic is the set of points for which two tangents of a given curve meet at a fixed angle (see 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 PQ can be considered as the orthoptic of two circles which are degenerated to the two points P and Q.

Orthoptic of a parabola

Any parabola can be transformed by a rigid motion (angles are not changed) into a parabola with equation y = ax^2. The slope at a point of the parabola is m = 2ax. Replacing x gives the parametric representation of the parabola with the tangent slope as parameter: \left(\tfrac{m}{2a},\tfrac{m^2}{4a} \right) ! . The tangent has the equation y=mx+n with the still unknown n, which can be determined by inserting the coordinates of the parabola point. One gets y=mx-\tfrac{m^2}{4a}; .

If a tangent contains the point (x0, y0), off the parabola, then the equation y_0 = m x_0 -\frac{m^2}{4a} \quad \rightarrow \quad m^2 - 4ax_0,m + 4ay_0 = 0 holds, which has two solutions m1 and m2 corresponding to the two tangents passing (x0, y0). The free term of a reduced quadratic equation is always the product of its solutions. Hence, if the tangents meet at (x0, y0) orthogonally, the following equations hold: m_1 m_2 = -1 = 4 a y_0 The last equation is equivalent to y_0 = -\frac{1}{4a}, , which is the equation of the directrix.

Orthoptic of an ellipse and hyperbola

Ellipse

Main article: Director circle

Let E:; \tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1 be the ellipse of consideration.

1. The tangents to the ellipse E at the vertices and co-vertices intersect at the 4 points (\pm a, \pm b), which lie on the desired orthoptic curve (the circle x^2+y^2 = a^2 + b^2).
2. The tangent at a point (u,v) of the ellipse E has the equation \tfrac{u}{a^2} x + \tfrac{v}{b^2} y = 1 (see tangent to an ellipse). If the point is not a vertex this equation can be solved for y: y = -\tfrac{b^2u}{a^2v};x; + ;\tfrac{b^2}{v}, .

Using the abbreviations m &= -\tfrac{b^2u}{a^2v},\ \color{red}n &= \color{red}\tfrac{b^2}{v} \end{align} |}} and the equation {\color{blue}\tfrac{u^2}{a^2} = 1 - \tfrac{v^2}{b^2} = 1-\tfrac{b^2}{n^2}} one gets: m^2 = \frac{b^4 u^2}{a^4 v^2} = \frac{1}{a^2} {\color{red}\frac{b^4}{v^2}} {\color{blue}\frac{u^2}{a^2}} = \frac{1}{a^2} {\color{red}n^2} {\color{blue}\left(1-\frac{b^2}{n^2}\right)} = \frac{n^2-b^2}{a^2}, . Hence and the equation of a non vertical tangent is y = m x \pm \sqrt{m^2 a^2 + b^2}. Solving relations for u,v and respecting leads to the slope depending parametric representation of the ellipse: (u,v) = \left(-\tfrac{ma^2}{\pm\sqrt{m^2a^2+b^2}};,;\tfrac{b^2}{\pm\sqrt{m^2a^2+b^2}}\right), . (For another proof: see .)

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. Eliminating 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 through (x_0,y_0). The constant term of a monic quadratic equation is always the product of its solutions. Hence, if the tangents meet at (x_0,y_0) orthogonally, the following equations hold: ::figure[src="https://upload.wikimedia.org/wikipedia/commons/b/b1/Orthoptic_locus_of_a_circle,_ellipses_and_hyperbolas.gif" caption="Orthoptics (red circles) of a circle, ellipses and hyperbolas"] ::

m_1 m_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, . From (1) and (2) one gets:

Hyperbola

The ellipse case can be adopted nearly exactly to the hyperbola case. The only changes to be made are to replace b^2 with -b^2 and to restrict m to . Therefore:

Orthoptic of an astroid

::figure[src="https://upload.wikimedia.org/wikipedia/commons/a/ae/Orthoptic-astroid.svg" caption="Orthoptic (purple) of an astroid"] ::

An astroid can be described by the parametric representation \mathbf c(t) = \left(\cos^3t, \sin^3t\right), \quad 0 \le t From the condition \mathbf \dot c(t) \cdot \mathbf \dot c(t+\alpha) = 0 one recognizes the distance α in parameter space at which an orthogonal tangent to ċ(t) appears. It turns out that the distance is independent of parameter t, namely . The equations of the (orthogonal) tangents at the points c(t) and c(t + ) are respectively: \begin{align} y &= -\tan t \left(x-\cos^3 t\right) + \sin^3 t, \ y &= \frac{1}{\tan t} \left(x+\sin^3 t\right) + \cos^3 t. \end{align} Their common point has coordinates: \begin{align} x &= \sin t \cos t \left(\sin t - \cos t\right), \ y &= \sin t \cos t \left(\sin t + \cos t\right). \end{align} This is simultaneously a parametric representation of the orthoptic.

Elimination of the parameter t yields the implicit representation 2\left(x^2+y^2\right)^3 - \left(x^2-y^2\right)^2 = 0. Introducing the new parameter one gets \begin{align} x &= \tfrac{1}{\sqrt{2}} \cos(2\varphi)\cos\varphi, \ y &= \tfrac{1}{\sqrt{2}} \cos(2\varphi)\sin\varphi. \end{align} (The proof uses the angle sum and difference identities.) Hence we get the polar representation r = \tfrac{1}{\sqrt{2}} \cos(2\varphi), \quad 0 \le \varphi of the orthoptic. Hence:

Isoptic of a parabola, an ellipse and a hyperbola

::figure[src="https://upload.wikimedia.org/wikipedia/commons/e/e0/Isoptic-80-parabola-s.svg" caption="Isoptics (purple) of a parabola for angles 80° and 100°"] ::

::figure[src="https://upload.wikimedia.org/wikipedia/commons/b/bd/Isoptic-ellipse-s.svg" caption="Isoptics (purple) of an ellipse for angles 80° and 100°"] ::

::figure[src="https://upload.wikimedia.org/wikipedia/commons/3/35/Isoptic-hyperbola-s.svg" caption="Isoptics (purple) of a hyperbola for angles 80° and 100°"] ::

Below the isotopics for angles α ≠ 90° are listed. They are called α-isoptics. For the proofs see below.

Equations of the isoptics

; Parabola: The α-isoptics of the parabola with equation are the branches of the hyperbola x^2-\tan^2\alpha\left(y+\frac{1}{4a}\right)^2-\frac{y}{a}=0. The branches of the hyperbola provide the isoptics for the two angles α and 180° − α (see picture).

; Ellipse: The α-isoptics of the ellipse with equation are the two parts of the degree-4 curve \left(x^2+y^2-a^2-b^2\right)^2 \tan^2\alpha = 4\left(a^2y^2 + b^2x^2 - a^2b^2\right) (see picture).

; Hyperbola: The α-isoptics of the hyperbola with the equation are the two parts of the degree-4 curve \left(x^2 + y^2 - a^2 + b^2\right)^2 \tan^2\alpha = 4 \left(a^2y^2 - b^2x^2 + a^2b^2\right).

Proofs

; Parabola: A parabola can be parametrized by the slope of its tangents : \mathbf c(m) = \left(\frac{m}{2a},\frac{m^2}{4a}\right), \quad m \in \R.

The tangent with slope m has the equation y=mx-\frac{m^2}{4a}.

The point (x0, y0) is on the tangent if and only if y_0 = m x_0 - \frac{m^2}{4a}.

This means the slopes m1, m2 of the two tangents containing (x0, y0) fulfil the quadratic equation m^2 - 4ax_0m + 4ay_0 = 0.

If the tangents meet at angle α or 180° − α, the equation \tan^2\alpha = \left(\frac{m_1-m_2}{1+m_1 m_2}\right)^2

must be fulfilled. Solving the quadratic equation for m, and inserting m1, m2 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 branches bear the two isoptics of the parabola for the two angles α and 180° − α.

; Ellipse: In the case of an ellipse one can adopt the idea for the orthoptic for 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 the case of a parabola, the quadratic equation has to be solved and the two solutions m1, m2 must be inserted into the equation \tan^2\alpha=\left(\frac{m_1-m_2}{1+m_1m_2}\right)^2.

Rearranging shows that the isoptics are parts of the degree-4 curve: \left(x_0^2+y_0^2-a^2-b^2\right)^2 \tan^2\alpha = 4\left(a^2y_0^2+b^2x_0^2-a^2b^2\right).

; Hyperbola: The solution for the case of a hyperbola can be adopted from the ellipse case by replacing b2 with −b2 (as in the case of the orthoptics, see above).

To visualize the isoptics, see implicit curve.

Notes

References

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