Orthoptic (geometry) explained

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.

Examples:

1. The orthoptic of a parabola is its directrix (proof: see below),

\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),\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 there are no orthogonal tangents, see below),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 (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 can be considered as the orthoptic of two circles which are degenerated to the two points and .

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 = 2 ax . Replacing gives the parametric representation of the parabola with the tangent slope as parameter: \left(\tfrac{m}{ 2 a},\tfrac{m ^{2}{4a}} \right) . The tangent has the equation y = mx + n with the still unknown, 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, off the parabola, then the equation$$y\_0\; =\; m\; x\_0\; -\backslash frac\; \backslash quad\; \backslash rightarrow\; \backslash quad\; m^2\; -\; 4ax\_0\backslash ,m\; +\; 4ay\_0\; =\; 0$$holds, which has two solutions and corresponding to the two tangents passing . The free term of a reduced quadratic equation is always the product of its solutions. Hence, if the tangents meet at orthogonally, the following equations hold: $$m\_1\; m\_2\; =\; -1\; =\; 4\; a\; y\_0$$The last equation is equivalent to$$y\_0\; =\; -\backslash frac\backslash ,,$$which is the equation of the directrix.

Orthoptic of an ellipse and hyperbola

Ellipse

See 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} ).

1. 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 = - \tfrac{b ^{2u}{a} ^{2v} x} + \tfrac{b ^{2}{v}} .

Using the abbreviationsand the equation

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

If a tangent contains the point

(x _{0,y} _{0)} , off the ellipse, then the equation$$y\_0\; =\; m\; x\_0\; \backslash pm\; \backslash sqrt$$holds. Eliminating the square root leads to$$m^2\; -\; \backslash fracm\; +\; \backslash frac\; =\; 0,$$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: $$m\_1\; m\_2\; =\; -1\; =\; \backslash frac$$The last equation is equivalent to$$x\_0^2+y\_0^2\; =\; a^2+b^2\backslash ,\; .$$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 to . Therefore:

Orthoptic of an astroid

An astroid can be described by the parametric representationFrom the condition$$\backslash mathbf\; \backslash dot\; c(t)\; \backslash cdot\; \backslash mathbf\; \backslash dot\; c(t+\backslash alpha)\; =\; 0$$one recognizes the distance in parameter space at which an orthogonal tangent to appears. It turns out that the distance is independent of parameter, namely . The equations of the (orthogonal) tangents at the points and are respectively:Their common point has coordinates:This is simultaneously a parametric representation of the orthoptic.

Elimination of the parameter yields the implicit representation $$2\backslash left(x^2+y^2\backslash right)^3\; -\; \backslash left(x^2-y^2\backslash right)^2\; =\; 0.$$Introducing the new parameter one gets(The proof uses the angle sum and difference identities.) Hence we get the polar representation of the orthoptic. Hence:

Isoptic of a parabola, an ellipse and a hyperbola

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

Equations of the isoptics

Parabola:The -isoptics of the parabola with equation are the branches of the hyperbola$$x^2-\backslash tan^2\backslash alpha\backslash left(y+\backslash frac\backslash right)^2-\backslash frac=0.$$The branches of the hyperbola provide the isoptics for the two angles and (see picture).

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

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

Proofs

Parabola:A parabola can be parametrized by the slope of its tangents :$$\backslash mathbf\; c(m)\; =\; \backslash left(\backslash frac,\backslash frac\backslash right),\; \backslash quad\; m\; \backslash in\; \backslash R.$$

The tangent with slope has the equation $$y=mx-\backslash frac.$$

The point is on the tangent if and only if$$y\_0\; =\; m\; x\_0\; -\; \backslash frac.$$

This means the slopes, of the two tangents containing fulfil the quadratic equation $$m^2\; -\; 4ax\_0m\; +\; 4ay\_0\; =\; 0.$$

If the tangents meet at angle or, the equation $$\backslash tan^2\backslash alpha\; =\; \backslash left(\backslash frac\backslash right)^2$$

must be fulfilled. Solving the quadratic equation for, and inserting, into the last equation, one gets$$x\_0^2-\backslash tan^2\backslash alpha\backslash left(y\_0+\backslash frac\backslash right)^2-\backslash frac\; =\; 0.$$

This is the equation of the hyperbola above. Its branches bear the two isoptics of the parabola for the two angles and .

Ellipse:In the case of an ellipse one can adopt the idea for the orthoptic for the quadratic equation$$m^2-\backslash fracm\; +\; \backslash frac\; =\; 0.$$

Now, as in the case of a parabola, the quadratic equation has to be solved and the two solutions, must be inserted into the equation$$\backslash tan^2\backslash alpha=\backslash left(\backslash frac\backslash right)^2.$$

Rearranging shows that the isoptics are parts of the degree-4 curve:$$\backslash left(x\_0^2+y\_0^2-a^2-b^2\backslash right)^2\; \backslash tan^2\backslash alpha\; =\; 4\backslash left(a^2y\_0^2+b^2x\_0^2-a^2b^2\backslash right).$$

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

To visualize the isoptics, see implicit curve.

External links

• Special Plane Curves.
• Mathworld
• Jan Wassenaar's Curves
• "Isoptic curve" at MathCurve
• "Orthoptic curve" at MathCurve

References

• Book: Lawrence, J. Dennis . A catalog of special plane curves . Dover Publications . 1972 . 0-486-60288-5 . 58–59 . registration .
• Boris. Odehnal . Equioptic Curves of Conic Sections . Journal for Geometry and Graphics. 14 . 2010 . 1 . 29–43 .
• Book: Schaal, Hermann . Lineare Algebra und Analytische Geometrie . III . Vieweg . 1977 . 3-528-03058-5 . 220.
• Book: Steiner, Jacob . Vorlesungen über synthetische Geometrie . B. G. Teubner . Leipzig . 1867 . Part 2, p. 186.
• Maurizio. Ternullo. Two new sets of ellipse related concyclic points. Journal of Geometry . 2009 . 94 . 1–2. 159–173. 10.1007/s00022-009-0005-7. 120011519.