# 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.
Ellipse and its orthoptic (purple)
Hyperbola with its orthoptic (purple)
Examples

The orthoptic of

1) a parabola is its directrix (proof: s. parabola),
2) an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the director circle $x^2+y^2=a^2+b^2$ (s. below),
3) a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \, a>b,$ is the circle $x^2+y^2=a^2-b^2$ (in case of $a\le b$ there are no orthogonal tangents, s. below),
4) an astroid $x^{2/3} + y^{2/3}=1$ is a quadrifolium with the polar equation
$r=\tfrac{1}{\sqrt{2}}\cos(2\varphi), \ 0\le \varphi <2\pi,$ (s. below).
Generalizations
1) An isoptic is the set of points for which two tangents of a given curve meet at a fixed angle (s. 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 $\overline{P_1P_2}$ can be considered as the orthoptic of two circles which are degenerated to the two points $P_1,P_2$.

Remark: In medicine there exists the term orthoptic, too.

## Orthoptic of an ellipse and hyperbola

### Ellipse

Main article: Director circle

The ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ can be represented by the unusual parametric representation [1]

• $\vec c_\pm(m)=\left(-\frac{ma^2}{\pm\sqrt{m^2a^2+b^2}},\frac{b^2}{\pm\sqrt{m^2a^2+b^2}}\right),\, m \in \R,$

where $m$ is the slope of the tangent at a point of the ellipse. $\vec c_+(m)$ describes the upper half and $\vec c_-(m)$ the lower half of the ellipse. The points $(\pm a,0)$) with tangents parallel to the y-axis are excluded. But this is no problem, because these tangents meet orthogonal the tangents parallel to the x-axis in the ellipse points $(0, \pm b)$ Hence the points $(\pm a, \pm b)$ are points of the desired orthoptic (circle $x^2+y^2=a^2+b^2$).

The tangent at point $\vec c_\pm(m)$ has the equation

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

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. Eliminatig 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 point $(x_0,y_0)$. The free term of a reduced quadratic equation is always the product of its solutions. Hence, if the tangents meet at point $(x_0,y_0)$ orthogonal, the follwing equations hold:

$m_1m_2=-1=\frac{y_0^2-b^2}{x_0^2-a^2}$

The last equation is equivalent to

$x_0^2+y_0^2=a^2+b^2$

This means:

• The intersection points of orthogonal tangents are points of the circle $x^2+y^2=a^2+b^2$.

### Hyperbola

The ellipse case can be adopted nearly literally to the hyperbola case. The only changes to be made are: 1) replace $b^2$ by $-b^2$ and 2) restrict $m$ by $|m|>b/a$. One gets:

• The intersection points of orthogonal tangents are points of the circle $x^2+y^2=a^2-b^2,\ a>b$.

## Orthoptic of an astroid

Orthoptic (purple) of an astroid

An astroid can be described by the parametric representation

$\vec c(t)=(\cos^3t,\sin^3t), \; 0\le t<2\pi$.

From the condition $\vec \dot c(t)\cdot\vec \dot c(t+\alpha)=0$ one recognizes, at what distance $\alpha$ in parameter space an orthogonal tangent to $\vec \dot c(t)$ appears. It turns out, that the distance is indepent of parameter $t$, namely $\alpha=\pm \tfrac{\pi}{2}$. The equations of the (orthogonal) tangents at the points $\vec c(t)$ and $\vec c(t+\tfrac{\pi}{2})$ are:

$y=-\tan t (x-\cos^3 t)+\sin^3t, \ y=\tfrac{1}{\tan t} (x+\sin^3 t)+\cos^3t$.

There common point has coordinates:

$x=\sin t\cos t(\sin t-\cos t)$,
$y=\sin t\cos t(\sin t+\cos t)$.

This is at the same time a parametric representation of the orthoptic.

Elimination of parameter $t$ yields the implicit representation

$2(x^2+y^2)^3-(x^2-y^2)^2=0.$

Introducing the new parameter $\varphi=t-\tfrac{5}{4}\pi$ one gets

$x=\tfrac{1}{\sqrt{2}}\cos(2\varphi)\,\cos\varphi, \ \ y=\tfrac{1}{\sqrt{2}}\cos(2\varphi)\,\sin\varphi$. (proof uses Angle sum and difference identities.)

Herefrom we get the polar representation

$r=\frac{1}{\sqrt{2}}\cos(2\varphi), \ 0\le \varphi <2\pi$

of the orthoptic. Hence

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

Isoptics (purple) of a parabola for angles 80 and 100 degree
Isoptics (purple) of an ellipse for angles 80 and 100 degree
Isoptics (purple) of a hyperbola for angles 80 and 100 degree

Below isotopics for an angle $\alpha\ne 90^\circ$ are listet. They are called $\alpha$-isoptics. For the proofs: s. below.

### Equations of the isoptics

parabola :

The $\alpha$-isoptics of the parabola with equation $y=ax^2$ are the arms of the hyperbola

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

The arms of the hyperbola provide the isoptics for the two angles $\alpha, 180^\circ\!-\!\alpha$ (s. picture).

Ellipse:

The $\alpha$-isoptics of the ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are the two parts of the curve of 4. degree

$\tan^2\alpha\;(x^2+y^2-a^2-b^2)^2=4(a^2y^2+b^2x^2-a^2b^2)$ (s. picture).
Hyperbola:

The $\alpha$-isoptics of the hyperbola with the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are the two parts of the curve of 4. degree

$\tan^2\alpha\;(x^2+y^2-a^2+b^2)^2=4(a^2y^2-b^2x^2+a^2b^2).$

### Proofs

Parabola:

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

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

The tangent with slope $m$ has the equation

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

Point $(x_0,y_0)$ is on the tangent, if

$y_0=mx_0-\frac{m^2}{4a}$

holds. That means, the slopes $m_1,m_2$ of the two tangents containing $(x_0,y_0)$ fulfill the quadratic equation

$m^2 - 4ax_0m + 4ay_0=0.$

If the tangents meet with angle $\alpha$ or $180^\circ -\alpha$, the equation

$\tan^2\alpha=\left(\frac{m_1-m_2}{1+m_1m_2}\right)^2$

has to be fulfilled. Solving the quadratic equation for $m,$ 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 arms bear the two isoptics of the parabola for the two angles $\alpha$ and $180^\circ-\alpha$.

Ellipse:

In case of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ one can adopt the idea for the orthoptic until the quadratic equation

$m^2-\frac{2x_0y_0}{x_0^2-a^2}m + \frac{y_0^2-b^2}{x_0^2-a^2}=0.$

Now, as in case of a parabola, the quadratic equation has to be solved and the two solutions $m_1,m_2$ to be inserted into the equation $\tan^2\alpha=\left(\tfrac{m_1-m_2}{1+m_1m_2}\right)^2$. Rearranging shows that the isoptics are parts of the curve of 4. degree:

$\tan^2\alpha\;(x_0^2+y_0^2-a^2-b^2)^2=4(a^2y_0^2+b^2x_0^2-a^2b^2).$
Hyperbola:

The solution for the case of a hyperbola can be adopted from the ellipse case by replacing $b^2$ by $-b^2$ (as in the case of the orthoptics, s. above).

Remark: For visualizing the isoptics see implicit curve.