# 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 ${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1}$ is the director circle ${\displaystyle x^{2}+y^{2}=a^{2}+b^{2}}$ (s. below),
3) a hyperbola ${\displaystyle {\frac {x^{2}}{a^{2}}}-{\frac {y^{2}}{b^{2}}}=1,\,a>b,}$ is the circle ${\displaystyle x^{2}+y^{2}=a^{2}-b^{2}}$ (in case of ${\displaystyle a\leq b}$ there are no orthogonal tangents, s. below),
4) an astroid ${\displaystyle x^{2/3}+y^{2/3}=1}$ is a quadrifolium with the polar equation
${\displaystyle r={\tfrac {1}{\sqrt {2}}}\cos(2\varphi ),\ 0\leq \varphi <2\pi ,}$ (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 ${\displaystyle {\overline {P_{1}P_{2}}}}$ can be considered as the orthoptic of two circles which are degenerated to the two points ${\displaystyle P_{1},P_{2}}$.

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

Orthoptics (red circles) of a circle, ellipses and hyperbolas

## Orthoptic of an ellipse and hyperbola

### Ellipse

Main article: Director circle

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

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

where ${\displaystyle m}$ is the slope of the tangent at a point of the ellipse. ${\displaystyle {\vec {c}}_{+}(m)}$ describes the upper half and ${\displaystyle {\vec {c}}_{-}(m)}$ the lower half of the ellipse. The points ${\displaystyle (\pm a,0)}$) 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 ${\displaystyle (0,\pm b)}$ Hence the points ${\displaystyle (\pm a,\pm b)}$ are points of the desired orthoptic (circle ${\displaystyle x^{2}+y^{2}=a^{2}+b^{2}}$).

The tangent at point ${\displaystyle {\vec {c}}_{\pm }(m)}$ has the equation

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

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. Eliminatig 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 ${\displaystyle m_{1},m_{2}}$ corresponding to the two tangents passing point ${\displaystyle (x_{0},y_{0})}$. The free term of a reduced quadratic equation is always the product of its solutions. Hence, if the tangents meet at point ${\displaystyle (x_{0},y_{0})}$ orthogonal, the following equations hold:

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

The last equation is equivalent to

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

This means:

• 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 literally to the hyperbola case. The only changes to be made are: 1) replace ${\displaystyle b^{2}}$ by ${\displaystyle -b^{2}}$ and 2) restrict ${\displaystyle m}$ by ${\displaystyle |m|>b/a}$. One gets:

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

## Orthoptic of an astroid

Orthoptic (purple) of an astroid

An astroid can be described by the parametric representation

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

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

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

Their common point has coordinates:

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

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

Elimination of parameter ${\displaystyle t}$ yields the implicit representation

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

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

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

Herefrom we get the polar representation

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

of the orthoptic. Hence

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

Isoptics (purple) of a parabola for angles 80 and 100 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 ${\displaystyle \alpha \neq 90^{\circ }}$ are listed. They are called ${\displaystyle \alpha }$-isoptics. For the proofs: s. below.

### Equations of the isoptics

parabola :

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

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

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

Ellipse:

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

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

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

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

### Proofs

Parabola:

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

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

The tangent with slope ${\displaystyle m}$ has the equation

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

Point ${\displaystyle (x_{0},y_{0})}$ is on the tangent, if

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

holds. That means, the slopes ${\displaystyle m_{1},m_{2}}$ of the two tangents containing ${\displaystyle (x_{0},y_{0})}$ fulfill the quadratic equation

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

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

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

has to be fulfilled. Solving the quadratic equation for ${\displaystyle m,}$ inserting ${\displaystyle m_{1},m_{2}}$ 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 arms bear the two isoptics of the parabola for the two angles ${\displaystyle \alpha }$ and ${\displaystyle 180^{\circ }-\alpha }$.

Ellipse:

In case of an ellipse ${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1}$ one can adopt the idea for the orthoptic until 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 case of a parabola, the quadratic equation has to be solved and the two solutions ${\displaystyle m_{1},m_{2}}$ to be inserted into the equation ${\displaystyle \tan ^{2}\alpha =\left({\tfrac {m_{1}-m_{2}}{1+m_{1}m_{2}}}\right)^{2}}$. Rearranging shows that the isoptics are parts of the curve of 4. degree:

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

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

Remark: For visualizing the isoptics see implicit curve.