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.

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Examples:

The orthoptic of a parabola is its directrix (proof: see below),
The orthoptic of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is the director circle $x^{2} + y^{2} = a^{2} + b^{2}$ (see below),
The orthoptic of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, a > b$ is the director circle $x^{2} + y^{2} = a^{2} - b^{2}$ (in case of Template:Math there are no orthogonal tangents, see below),
The orthoptic of an astroid $x^{2 / 3} + y^{2 / 3} = 1$ is a quadrifolium with the polar equation $r = \frac{1}{\sqrt{2}} \cos (2 φ), 0 \leq φ < 2 π$ (see below).

Generalizations:

An isoptic is the set of points for which two tangents of a given curve meet at a fixed angle (see below).
An isoptic of two plane curves is the set of points for which two tangents meet at a fixed angle.
Thales' theorem on a chord Template:Mvar can be considered as the orthoptic of two circles which are degenerated to the two points Template:Mvar and Template:Mvar.

Orthoptic of a parabola

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

If a tangent contains the point Template:Math, off the parabola, then the equation $y_{0} = m x_{0} - \frac{m^{2}}{4 a} \to m^{2} - 4 a x_{0} m + 4 a y_{0} = 0$ holds, which has two solutions Template:Math and Template:Math corresponding to the two tangents passing Template:Math. The free term of a reduced quadratic equation is always the product of its solutions. Hence, if the tangents meet at Template:Math orthogonally, the following equations hold: $m_{1} m_{2} = - 1 = 4 a y_{0}$ The last equation is equivalent to $y_{0} = - \frac{1}{4 a},$ which is the equation of the directrix.

Orthoptic of an ellipse and hyperbola

Ellipse

Template:Main Let $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ be the ellipse of consideration.

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 $x^{2} + y^{2} = a^{2} + b^{2}$ ).
The tangent at a point $(u, v)$ of the ellipse $E$ has the equation $\frac{u}{a^{2}} x + \frac{v}{b^{2}} y = 1$ (see tangent to an ellipse). If the point is not a vertex this equation can be solved for Template:Mvar: $y = - \frac{b^{2} u}{a^{2} v} x + \frac{b^{2}}{v} .$

Using the abbreviations Template:NumBlk and the equation $\frac{u^{2}}{a^{2}} = 1 - \frac{v^{2}}{b^{2}} = 1 - \frac{b^{2}}{n^{2}}$ one gets: $m^{2} = \frac{b^{4} u^{2}}{a^{4} v^{2}} = \frac{1}{a^{2}} \frac{b^{4}}{v^{2}} \frac{u^{2}}{a^{2}} = \frac{1}{a^{2}} n^{2} (1 - \frac{b^{2}}{n^{2}}) = \frac{n^{2} - b^{2}}{a^{2}} .$ Hence Template:NumBlk and the equation of a non vertical tangent is $y = m x \pm \sqrt{m^{2} a^{2} + b^{2}} .$ Solving relations Template:EquationNote for $u, v$ and respecting Template:EquationNote leads to the slope depending parametric representation of the ellipse: $(u, v) = (- \frac{m a^{2}}{\pm \sqrt{m^{2} a^{2} + b^{2}}}, \frac{b^{2}}{\pm \sqrt{m^{2} a^{2} + b^{2}}}) .$ (For another proof: see Template:Slink.)

If a tangent contains the point $(x_{0}, y_{0})$ , off the ellipse, then the equation $y_{0} = m x_{0} \pm \sqrt{m^{2} a^{2} + b^{2}}$ holds. Eliminating the square root leads to $m^{2} - \frac{2 x_{0} y_{0}}{x_{0}^{2} - a^{2}} m + \frac{y_{0}^{2} - b^{2}}{x_{0}^{2} - a^{2}} = 0,$ 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:

$m_{1} m_{2} = - 1 = \frac{y_{0}^{2} - b^{2}}{x_{0}^{2} - a^{2}}$ The last equation is equivalent to $x_{0}^{2} + y_{0}^{2} = a^{2} + b^{2} .$ From (1) and (2) one gets: Template:Block indent

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 Template:Mvar to Template:Math. Therefore: Template:Block indent

Orthoptic of an astroid

An astroid can be described by the parametric representation $𝐜 (t) = (\cos^{3} t, \sin^{3} t), 0 \leq t < 2 π .$ From the condition $\dot{𝐜} (t) \cdot \dot{𝐜} (t + α) = 0$ one recognizes the distance Template:Mvar in parameter space at which an orthogonal tangent to Template:Math appears. It turns out that the distance is independent of parameter Template:Mvar, namely Template:Math. The equations of the (orthogonal) tangents at the points Template:Math and Template:Math are respectively: $\begin{matrix} y & = - \tan t (x - \cos^{3} t) + \sin^{3} t, \\ y & = \frac{1}{\tan t} (x + \sin^{3} t) + \cos^{3} t . \end{matrix}$ Their common point has coordinates: $\begin{matrix} x & = \sin t \cos t (\sin t - \cos t), \\ y & = \sin t \cos t (\sin t + \cos t) . \end{matrix}$ This is simultaneously a parametric representation of the orthoptic.

Elimination of the parameter Template:Mvar yields the implicit representation $2 {(x^{2} + y^{2})}^{3} - {(x^{2} - y^{2})}^{2} = 0 .$ Introducing the new parameter Template:Math one gets $\begin{matrix} x & = \frac{1}{\sqrt{2}} \cos (2 φ) \cos φ, \\ y & = \frac{1}{\sqrt{2}} \cos (2 φ) \sin φ . \end{matrix}$ (The proof uses the angle sum and difference identities.) Hence we get the polar representation $r = \frac{1}{\sqrt{2}} \cos (2 φ), 0 \leq φ < 2 π$ of the orthoptic. Hence: Template:Block indent

Isoptic of a parabola, an ellipse and a hyperbola

Isoptics (purple) of a parabola for angles 80° and 100°

Isoptics (purple) of an ellipse for angles 80° and 100°

Isoptics (purple) of a hyperbola for angles 80° and 100°

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

Equations of the isoptics

Parabola:

The Template:Mvar-isoptics of the parabola with equation Template:Math are the branches of the hyperbola $x^{2} - \tan^{2} α {(y + \frac{1}{4 a})}^{2} - \frac{y}{a} = 0 .$ The branches of the hyperbola provide the isoptics for the two angles Template:Mvar and Template:Mvar (see picture).

Ellipse:

The Template:Mvar-isoptics of the ellipse with equation Template:Math are the two parts of the degree-4 curve ${(x^{2} + y^{2} - a^{2} - b^{2})}^{2} \tan^{2} α = 4 (a^{2} y^{2} + b^{2} x^{2} - a^{2} b^{2})$ (see picture).

Hyperbola:

The Template:Mvar-isoptics of the hyperbola with the equation Template:Math are the two parts of the degree-4 curve ${(x^{2} + y^{2} - a^{2} + b^{2})}^{2} \tan^{2} α = 4 (a^{2} y^{2} - b^{2} x^{2} + a^{2} b^{2}) .$

Proofs

Parabola:

A parabola Template:Math can be parametrized by the slope of its tangents Template:Math: $𝐜 (m) = (\frac{m}{2 a}, \frac{m^{2}}{4 a}), m \in ℝ .$

The tangent with slope Template:Mvar has the equation $y = m x - \frac{m^{2}}{4 a} .$

The point Template:Math is on the tangent if and only if $y_{0} = m x_{0} - \frac{m^{2}}{4 a} .$

This means the slopes Template:Math, Template:Math of the two tangents containing Template:Math fulfil the quadratic equation $m^{2} - 4 a x_{0} m + 4 a y_{0} = 0 .$

If the tangents meet at angle Template:Mvar or Template:Math, the equation $\tan^{2} α = {(\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}})}^{2}$

must be fulfilled. Solving the quadratic equation for Template:Mvar, and inserting Template:Math, Template:Math into the last equation, one gets $x_{0}^{2} - \tan^{2} α {(y_{0} + \frac{1}{4 a})}^{2} - \frac{y_{0}}{a} = 0 .$

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

Ellipse:

In the case of an ellipse Template:Math one can adopt the idea for the orthoptic for the quadratic equation $m^{2} - \frac{2 x_{0} y_{0}}{x_{0}^{2} - a^{2}} m + \frac{y_{0}^{2} - b^{2}}{x_{0}^{2} - a^{2}} = 0 .$

Now, as in the case of a parabola, the quadratic equation has to be solved and the two solutions Template:Math, Template:Math must be inserted into the equation $\tan^{2} α = {(\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}})}^{2} .$

Rearranging shows that the isoptics are parts of the degree-4 curve: ${(x_{0}^{2} + y_{0}^{2} - a^{2} - b^{2})}^{2} \tan^{2} α = 4 (a^{2} y_{0}^{2} + b^{2} x_{0}^{2} - a^{2} b^{2}) .$

Hyperbola:

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

To visualize the isoptics, see implicit curve.

External links

Template:Commons category

Notes

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References

Template:Differential transforms of plane curves

Orthoptic (geometry)

Contents

Orthoptic of a parabola