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),
2. The orthoptic of an ellipse ${\displaystyle {\tfrac {x^{2)){a^{2))}+{\tfrac {y^{2)){b^{2))}=1}$ is the director circle ${\displaystyle x^{2}+y^{2}=a^{2}+b^{2))$ (see below),
3. The orthoptic of a hyperbola ${\displaystyle {\tfrac {x^{2)){a^{2))}-{\tfrac {y^{2)){b^{2))}=1,\ a>b}$ is the director circle ${\displaystyle x^{2}+y^{2}=a^{2}-b^{2))$ (in case of a ≤ b there are no orthogonal tangents, see below),
4. The orthoptic of an astroid ${\displaystyle x^{2/3}+y^{2/3}=1}$ is a quadrifolium with the polar equation ${\displaystyle r={\tfrac {1}{\sqrt {2))}\cos(2\varphi ),\ 0\leq \varphi <2\pi }$ (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 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 ${\displaystyle y=ax^{2))$. The slope at a point of the parabola is ${\displaystyle m=2ax}$. Replacing x gives the parametric representation of the parabola with the tangent slope as parameter: ${\displaystyle \left({\tfrac {m}{2a)),{\tfrac {m^{2)){4a))\right)\!.}$ The tangent has the equation ${\displaystyle y=mx+n}$ with the still unknown n, which can be determined by inserting the coordinates of the parabola point. One gets ${\displaystyle y=mx-{\tfrac {m^{2)){4a))\;.}$

If a tangent contains the point (x₀, y₀), off the parabola, then the equation

$${\displaystyle y_{0}=mx_{0}-{\frac {m^{2)){4a))\quad \rightarrow \quad m^{2}-4ax_{0}\,m+4ay_{0}=0}$$

$${\displaystyle m_{1}m_{2}=-1=4ay_{0))$$

$${\displaystyle y_{0}=-{\frac {1}{4a))\,,}$$

Orthoptic of an ellipse and hyperbola

Ellipse

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

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

Using the abbreviations

$${\displaystyle {\begin{aligned}m&=-{\tfrac {b^{2}u}{a^{2}v)),\\\color {red}n&=\color {red}{\tfrac {b^{2)){v))\end{aligned))}$$

(I)

and the equation ${\displaystyle {\color {blue}{\tfrac {u^{2)){a^{2))}=1-{\tfrac {v^{2)){b^{2))}=1-{\tfrac {b^{2)){n^{2))))}$ one gets:

$${\displaystyle m^{2}={\frac {b^{4}u^{2)){a^{4}v^{2))}={\frac {1}{a^{2))}{\color {red}{\frac {b^{4)){v^{2)))){\color {blue}{\frac {u^{2)){a^{2))))={\frac {1}{a^{2))}{\color {red}n^{2)){\color {blue}\left(1-{\frac {b^{2)){n^{2))}\right)}={\frac {n^{2}-b^{2)){a^{2))}\,.}$$

$${\displaystyle n=\pm {\sqrt {m^{2}a^{2}+b^{2))))$$

(II)

and the equation of a non vertical tangent is

$${\displaystyle y=mx\pm {\sqrt {m^{2}a^{2}+b^{2))}.}$$

${\displaystyle u,v}$

$${\displaystyle (u,v)=\left(-{\tfrac {ma^{2)){\pm {\sqrt {m^{2}a^{2}+b^{2))))}\;,\;{\tfrac {b^{2)){\pm {\sqrt {m^{2}a^{2}+b^{2))))}\right)\,.}$$

If a tangent contains the point ${\displaystyle (x_{0},y_{0})}$, off the ellipse, then the equation

$${\displaystyle y_{0}=mx_{0}\pm {\sqrt {m^{2}a^{2}+b^{2))))$$

$${\displaystyle m^{2}-{\frac {2x_{0}y_{0)){x_{0}^{2}-a^{2))}m+{\frac {y_{0}^{2}-b^{2)){x_{0}^{2}-a^{2))}=0,}$$

${\displaystyle m_{1},m_{2))$

${\displaystyle (x_{0},y_{0})}$

${\displaystyle (x_{0},y_{0})}$

$${\displaystyle m_{1}m_{2}=-1={\frac {y_{0}^{2}-b^{2)){x_{0}^{2}-a^{2))))$$

$${\displaystyle x_{0}^{2}+y_{0}^{2}=a^{2}+b^{2}\,.}$$

The intersection points of orthogonal tangents are points of the circle ${\displaystyle x^{2}+y^{2}=a^{2}+b^{2))$.

Hyperbola

The ellipse case can be adopted nearly exactly to the hyperbola case. The only changes to be made are to replace ${\displaystyle b^{2))$ with ${\displaystyle -b^{2))$ and to restrict m to |m| > b/a. Therefore:

The intersection points of orthogonal tangents are points of the circle ${\displaystyle x^{2}+y^{2}=a^{2}-b^{2))$, where a > b.

Orthoptic of an astroid

An astroid can be described by the parametric representation

$${\displaystyle \mathbf {c} (t)=\left(\cos ^{3}t,\sin ^{3}t\right),\quad 0\leq t<2\pi .}$$

$${\displaystyle \mathbf {\dot {c)) (t)\cdot \mathbf {\dot {c)) (t+\alpha )=0}$$

$${\displaystyle {\begin{aligned}y&=-\tan t\left(x-\cos ^{3}t\right)+\sin ^{3}t,\\y&={\frac {1}{\tan t))\left(x+\sin ^{3}t\right)+\cos ^{3}t.\end{aligned))}$$

$${\displaystyle {\begin{aligned}x&=\sin t\cos t\left(\sin t-\cos t\right),\\y&=\sin t\cos t\left(\sin t+\cos t\right).\end{aligned))}$$

Elimination of the parameter t yields the implicit representation

$${\displaystyle 2\left(x^{2}+y^{2}\right)^{3}-\left(x^{2}-y^{2}\right)^{2}=0.}$$

$${\displaystyle {\begin{aligned}x&={\tfrac {1}{\sqrt {2))}\cos(2\varphi )\cos \varphi ,\\y&={\tfrac {1}{\sqrt {2))}\cos(2\varphi )\sin \varphi .\end{aligned))}$$

$${\displaystyle r={\tfrac {1}{\sqrt {2))}\cos(2\varphi ),\quad 0\leq \varphi <2\pi }$$

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 = ax² are the branches of the hyperbola

$${\displaystyle x^{2}-\tan ^{2}\alpha \left(y+{\frac {1}{4a))\right)^{2}-{\frac {y}{a))=0.}$$

Ellipse:

The α-isoptics of the ellipse with equation x²/a² + y²/b² = 1 are the two parts of the degree-4 curve

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

Hyperbola:

The α-isoptics of the hyperbola with the equation x²/a² − y²/b² = 1 are the two parts of the degree-4 curve

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

Proofs

Parabola:

A parabola y = ax² can be parametrized by the slope of its tangents m = 2ax:

$${\displaystyle \mathbf {c} (m)=\left({\frac {m}{2a)),{\frac {m^{2)){4a))\right),\quad m\in \mathbb {R} .}$$

The tangent with slope m has the equation

$${\displaystyle y=mx-{\frac {m^{2)){4a)).}$$

The point (x₀, y₀) is on the tangent if and only if

$${\displaystyle y_{0}=mx_{0}-{\frac {m^{2)){4a)).}$$

This means the slopes m₁, m₂ of the two tangents containing (x₀, y₀) fulfil the quadratic equation

$${\displaystyle m^{2}-4ax_{0}m+4ay_{0}=0.}$$

If the tangents meet at angle α or 180° − α, the equation

$${\displaystyle \tan ^{2}\alpha =\left({\frac {m_{1}-m_{2)){1+m_{1}m_{2))}\right)^{2))$$

must be fulfilled. Solving the quadratic equation for m, and inserting m₁, m₂ into the last equation, one gets

$${\displaystyle x_{0}^{2}-\tan ^{2}\alpha \left(y_{0}+{\frac {1}{4a))\right)^{2}-{\frac {y_{0)){a))=0.}$$

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 x²/a² + y²/b² = 1 one can adopt the idea for the orthoptic for the quadratic equation

$${\displaystyle m^{2}-{\frac {2x_{0}y_{0)){x_{0}^{2}-a^{2))}m+{\frac {y_{0}^{2}-b^{2)){x_{0}^{2}-a^{2))}=0.}$$

Now, as in the case of a parabola, the quadratic equation has to be solved and the two solutions m₁, m₂ must be inserted into the equation

$${\displaystyle \tan ^{2}\alpha =\left({\frac {m_{1}-m_{2)){1+m_{1}m_{2))}\right)^{2}.}$$

Rearranging shows that the isoptics are parts of the degree-4 curve:

$${\displaystyle \left(x_{0}^{2}+y_{0}^{2}-a^{2}-b^{2}\right)^{2}\tan ^{2}\alpha =4\left(a^{2}y_{0}^{2}+b^{2}x_{0}^{2}-a^{2}b^{2}\right).}$$

Hyperbola:

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

To visualize the isoptics, see implicit curve.