# 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.
Examples:
* The orthoptic of a parabola is its directrix (proof: see below),
* 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),
* 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),
* The orthoptic of an astroid $x^{2/3}+y^{2/3}=1$ is a quadrifolium with the polar equation $r={\tfrac {1}{\sqrt {2}}}\cos(2\varphi ),\ 0\leq \varphi <2\pi$ (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 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 (x$_0$, y$_0$), off the parabola, then the equation $y_{0}=mx_{0}-{\frac {m^{2}}{4a}}\quad \rightarrow \quad m^{2}-4ax_{0}\,m+4ay_{0}=0$ 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: $m_{1}m_{2}=-1=4ay_{0}$ 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

Let $E:\;{\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1$ 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 ${\tfrac {u}{a^{2}}}x+{\tfrac {v}{b^{2}}}y=1$ (see tangent to an ellipse). If the point is not a vertex this equation can be solved for y: $y=-{\tfrac {b^{2}u}{a^{2}v}}\;x\;+\;{\tfrac {b^{2}}{v}}\,.$
Using the abbreviations
| ${\begin{aligned}m&=-{\tfrac {b^{2}u}{a^{2}v}},\\\color {red}n&=\color {red}{\tfrac {b^{2}}{v}}\end{aligned}}$ | (I) |
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
| $n=\pm {\sqrt {m^{2}a^{2}+b^{2}}}$ | (II) |
and the equation of a non vertical tangent is $y=mx\pm {\sqrt {m^{2}a^{2}+b^{2}}}.$ Solving relations (I) for $u,v$ and respecting (II) leads to the slope depending parametric representation of the ellipse: $(u,v)=\left(-{\tfrac {ma^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\;,\;{\tfrac {b^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\right)\,.$ (For another proof: see Ellipse § Parametric representation.)
If a tangent contains the point $(x_{0},y_{0})$, off the ellipse, then the equation $y_{0}=mx_{0}\pm {\sqrt {m^{2}a^{2}+b^{2}}}$ holds. Eliminating the square root leads to $m^{2}-{\frac {2x_{0}y_{0}}{x_{0}^{2}-a^{2}}}m+{\frac {y_{0}^{2}-b^{2}}{x_{0}^{2}-a^{2}}}=0,$ 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:
$x^{2}+y^{2}=a^{2}+b^{2}$

### 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 |m| > ⁠b/a⁠. Therefore:
$x^{2}+y^{2}=a^{2}-b^{2}$

## Orthoptic of an astroid

An astroid can be described by the parametric representation $\mathbf {c} (t)=\left(\cos ^{3}t,\sin ^{3}t\right),\quad 0\leq t<2\pi .$ 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 α = ± ⁠π/2⁠. The equations of the (orthogonal) tangents at the points c(t) and c(t + ⁠π/2⁠) are respectively: ${\begin{aligned}y&=-\tan t\left(x-\cos ^{3}t\right)+\sin ^{3}t,\\y&={\frac {1}{\tan t}}\left(x+\sin ^{3}t\right)+\cos ^{3}t.\end{aligned}}$ Their common point has coordinates: ${\begin{aligned}x&=\sin t\cos t\left(\sin t-\cos t\right),\\y&=\sin t\cos t\left(\sin t+\cos t\right).\end{aligned}}$ 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 φ = t − ⁠5π/4⁠ one gets ${\begin{aligned}x&={\tfrac {1}{\sqrt {2}}}\cos(2\varphi )\cos \varphi ,\\y&={\tfrac {1}{\sqrt {2}}}\cos(2\varphi )\sin \varphi .\end{aligned}}$ (The proof uses the angle sum and difference identities.) Hence we get the polar representation $r={\tfrac {1}{\sqrt {2}}}\cos(2\varphi ),\quad 0\leq \varphi <2\pi$ of the orthoptic. Hence:

## Isoptic of a parabola, an ellipse and a hyperbola

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 y = ax$^2$ 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 ⁠x$^2$/a$^2$⁠ + ⁠y$^2$/b$^2$⁠ = 1 are the two parts of the degree-4 curve $\left(x^{2}+y^{2}-a^{2}-b^{2}\right)^{2}\tan ^{2}\alpha =4\left(a^{2}y^{2}+b^{2}x^{2}-a^{2}b^{2}\right)$ (see picture).

#### Hyperbola

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 $\left(x^{2}+y^{2}-a^{2}+b^{2}\right)^{2}\tan ^{2}\alpha =4\left(a^{2}y^{2}-b^{2}x^{2}+a^{2}b^{2}\right).$

### Proofs
#### Parabola

A parabola y = ax$^2$ can be parametrized by the slope of its tangents m = 2ax: $\mathbf {c} (m)=\left({\frac {m}{2a}},{\frac {m^{2}}{4a}}\right),\quad m\in \mathbb {R} .$
The tangent with slope m has the equation $y=mx-{\frac {m^{2}}{4a}}.$
The point (x$_0$, y$_0$) is on the tangent if and only if $y_{0}=mx_{0}-{\frac {m^{2}}{4a}}.$
This means the slopes m$_1$, m$_2$ of the two tangents containing (x$_0$, y$_0$) fulfil the quadratic equation $m^{2}-4ax_{0}m+4ay_{0}=0.$
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 m$_1$, m$_2$ 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 ⁠x$^2$/a$^2$⁠ + ⁠y$^2$/b$^2$⁠ = 1 one can adopt the idea for the orthoptic for the quadratic equation $m^{2}-{\frac {2x_{0}y_{0}}{x_{0}^{2}-a^{2}}}m+{\frac {y_{0}^{2}-b^{2}}{x_{0}^{2}-a^{2}}}=0.$
Now, as in the case of a parabola, the quadratic equation has to be solved and the two solutions m$_1$, m$_2$ must be inserted into the equation $\tan ^{2}\alpha =\left({\frac {m_{1}-m_{2}}{1+m_{1}m_{2}}}\right)^{2}.$
Rearranging shows that the isoptics are parts of the degree-4 curve: $\left(x_{0}^{2}+y_{0}^{2}-a^{2}-b^{2}\right)^{2}\tan ^{2}\alpha =4\left(a^{2}y_{0}^{2}+b^{2}x_{0}^{2}-a^{2}b^{2}\right).$

#### Hyperbola

The solution for the case of a hyperbola can be adopted from the ellipse case by replacing b$^2$ with −b$^2$ (as in the case of the orthoptics, see above).
To visualize the isoptics, see implicit curve.