Orthoptic (geometry)

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.

[[File:Parabel-orthop.svg|right|thumb|

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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 $$r=\tfrac{1}{\sqrt{2}}\cos(2\varphi), \ 0\le \varphi < 2\pi$$ (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 $xy$ can be considered as the orthoptic of two circles which are degenerated to the two points $PQ$ and $P$.

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 $Q$ 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 $x$, 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 $(x_{0}, y_{0})$, 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 $m_{1}$ and $m_{2}$ corresponding to the two tangents passing $(x_{0}, y_{0})$. The free term of a reduced quadratic equation is always the product of its solutions. Hence, if the tangents meet at $(x_{0}, y_{0})$ 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.

Ellipse
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 $n$: $$ y = -\tfrac{b^2u}{a^2v}\;x\; + \;\tfrac{b^2}{v}\, .$$

Using the abbreviations

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 $y$ 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: $$m_1 m_2 = -1 = \frac{y_0^2 - b^2}{x_0^2 - a^2}$$ 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 $$ to $|m| > b⁄a$. Therefore:

Orthoptic of an astroid
An astroid can be described by the parametric representation $$\mathbf c(t) = \left(\cos^3t, \sin^3t\right), \quad 0 \le t < 2\pi.$$ From the condition $$\mathbf \dot c(t) \cdot \mathbf \dot c(t+\alpha) = 0$$ one recognizes the distance $$ in parameter space at which an orthogonal tangent to $a > b$ appears. It turns out that the distance is independent of parameter $$, namely $ċ(t)$. The equations of the (orthogonal) tangents at the points $α = ± π⁄2$ 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 $m$ yields the implicit representation $$2\left(x^2+y^2\right)^3 - \left(x^2-y^2\right)^2 = 0.$$ Introducing the new parameter $c(t + π⁄2)$ 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 $$r = \tfrac{1}{\sqrt{2}} \cos(2\varphi), \quad 0 \le \varphi < 2\pi$$ of the orthoptic. Hence:

Isoptic of a parabola, an ellipse and a hyperbola
Below the isotopics for angles $φ = t − 5π⁄4$ are listed. They are called $α$-isoptics. For the proofs see below.

Equations of the isoptics
The $t$-isoptics of the parabola with equation $α ≠ 90°$ 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 $t$ and $α$ (see picture).
 * Parabola :

The $α$-isoptics of the ellipse with equation $y = ax^{2}$ 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).
 * Ellipse :

The $α$-isoptics of the hyperbola with the equation $x^{2}⁄a^{2} + y^{2}⁄b^{2} = 1$ are the two parts of the degree-4 curve $$\left(x^2 + y^2 - a^2 + b^2\right)^2 \tan^2\alpha = 4 \left(a^2y^2 - b^2x^2 + a^2b^2\right).$$
 * Hyperbola :

Proofs
A parabola $x^{2}⁄a^{2} − y^{2}⁄b^{2} = 1$ can be parametrized by the slope of its tangents $y = ax^{2}$: $$\mathbf c(m) = \left(\frac{m}{2a},\frac{m^2}{4a}\right), \quad m \in \R.$$
 * Parabola :

The tangent with slope $180° − α$ has the equation $$y=mx-\frac{m^2}{4a}.$$

The point $m = 2ax$ is on the tangent if and only if $$y_0 = m x_0 - \frac{m^2}{4a}.$$

This means the slopes $(x_{0}, y_{0})$, $m_{1}$ of the two tangents containing $m_{2}$ fulfil the quadratic equation $$m^2 - 4ax_0m + 4ay_0 = 0.$$

If the tangents meet at angle $α$ or $(x_{0}, y_{0})$, 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 $α$, and inserting $180° − α$, $m_{1}$ 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 $m$ and $m_{2}$.

In the case of an ellipse $180° − α$ one can adopt the idea for the orthoptic for the quadratic equation $$m^2-\frac{2x_0y_0}{x_0^2-a^2}m + \frac{y_0^2-b^2}{x_0^2-a^2} = 0.$$
 * Ellipse :

Now, as in the case of a parabola, the quadratic equation has to be solved and the two solutions $x^{2}⁄a^{2} + y^{2}⁄b^{2} = 1$, $m_{1}$ 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: $$\left(x_0^2+y_0^2-a^2-b^2\right)^2 \tan^2\alpha = 4\left(a^2y_0^2+b^2x_0^2-a^2b^2\right).$$

The solution for the case of a hyperbola can be adopted from the ellipse case by replacing $m_{2}$ with $b^{2}$ (as in the case of the orthoptics, see above).
 * Hyperbola :

To visualize the isoptics, see implicit curve.