# Orthoptic (geometry)

(Redirected from Isoptic)

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.

The orthoptic of a parabola is its directrix (purple).
Ellipse and its orthoptic (purple)
Hyperbola with its orthoptic (purple)

Examples:

1. The orthoptic of a parabola is its directrix (proof: see below),
2. The orthoptic of an ellipse x2/a2 + y2/b2 = 1 is the director circle x2 + y2 = a2 + b2 (see below),
3. The orthoptic of a hyperbola x2/a2y2/b2 = 1, a > b, is the circle x2 + y2 = a2b2 (in case of ab there are no orthogonal tangents, see below),
4. The orthoptic of an astroid x23 + y23 = 1 is a quadrifolium with the polar equation
${\displaystyle r={\tfrac {1}{\sqrt {2}}}\cos(2\varphi ),\quad 0\leq \varphi <2\pi }$
(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 ${\displaystyle x}$ gives the parametric representation of the parabola with the tangent slope as parameter: ${\displaystyle \ ({\tfrac {m}{2a}},{\tfrac {m^{2}}{4a}})\;.}$ The tangent has the equation ${\displaystyle y=mx+n}$ with the still unknown ${\displaystyle n}$, 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

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

holds, which has two solutions m1 and m2 corresponding to the two tangents passing (x0, y0). The free term of a reduced quadratic equation is always the product of its solutions. Hence, if the tangents meet at (x0, y0) orthogonally, the following equations hold:

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

The last equation is equivalent to

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

which is the equation of the directrix.

## 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 of ellipse ${\displaystyle E}$ at neighbored vertices intersect at one of the 4 points ${\displaystyle (\pm a,\pm b)}$, which lie on the desired orthoptic curve (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 ${\displaystyle {\tfrac {u}{a^{2}}}x+{\tfrac {v}{b^{2}}}y=1}$ (s. Ellipse). If the point is not a vertex this equation can be solved: ${\displaystyle \ y=-{\tfrac {b^{2}u}{a^{2}v}}\;x\;+\;{\tfrac {b^{2}}{v}}\;.}$

Using the abbreviations ${\displaystyle (I)\;m=-{\tfrac {b^{2}u}{a^{2}v}},\;{\color {red}n={\tfrac {b^{2}}{v}}}\;}$ 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:

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

Hence ${\displaystyle \ (II)\;n=\pm {\sqrt {m^{2}a^{2}+b^{2}}}}$ and the equation of a non vertical tangent is

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

Solving relations ${\displaystyle (I)}$ for ${\displaystyle u,v}$ and respecting ${\displaystyle (II)}$ leads to the slope depending parametric representation of the ellipse:

${\displaystyle (u,v)=(-{\tfrac {ma^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\;,\;{\tfrac {b^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}})\ .\ }$ (For an other proof: see Ellipse.)

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

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

holds. Eliminating the square root leads to

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

which has two solutions ${\displaystyle (x_{0},y_{0})}$ corresponding to the two tangents passing ${\displaystyle (x_{0},y_{0})}$. The free term of a reduced quadratic equation is always the product of its solutions. Hence, if the tangents meet at ${\displaystyle (x_{0},y_{0})}$ orthogonally, the following equations hold:

Orthoptics (red circles) of a circle, ellipses and hyperbolas
${\displaystyle m_{1}m_{2}=-1={\frac {y_{0}^{2}-b^{2}}{x_{0}^{2}-a^{2}}}}$

The last equation is equivalent to

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

From (1) and (2) one gets:

• 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

Orthoptic (purple) of an astroid

An astroid can be described by the parametric representation

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

From the condition

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

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:

{\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}}}

Their common point has coordinates:

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

This is simultaneously a parametric representation of the orthoptic.

Elimination of the parameter t yields the implicit representation

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

Introducing the new parameter φ = t/4 one gets

{\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}}}

(The proof uses the angle sum and difference identities.) Hence we get the polar representation

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

of the orthoptic. Hence:

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

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 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

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

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 x2/a2 + y2/b2 = 1 are the two parts of the degree-4 curve

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

(see picture).

Hyperbola:

The α-isoptics of the hyperbola with the equation x2/a2y2/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:

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

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.