All points for which two tangents of a curve intersect at 90° angles
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.
Orthoptic of the hyperbola (its director circle)
Examples:
The orthoptic of a parabola is its directrix (proof: see below ),
The orthoptic of an ellipse
x
2
a
2
+
y
2
b
2
=
1
{\displaystyle {\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1}
is the director circle
x
2
+
y
2
=
a
2
+
b
2
{\displaystyle x^{2}+y^{2}=a^{2}+b^{2}}
(see below ),
The orthoptic of a hyperbola
x
2
a
2
−
y
2
b
2
=
1
,
a
>
b
{\displaystyle {\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
{\displaystyle 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
{\displaystyle x^{2/3}+y^{2/3}=1}
is a quadrifolium with the polar equation
r
=
1
2
cos
(
2
φ
)
,
0
≤
φ
<
2
π
{\displaystyle 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 [ edit ]
Any parabola can be transformed by a rigid motion (angles are not changed) into a parabola with equation
y
=
a
x
2
{\displaystyle y=ax^{2}}
. The slope at a point of the parabola is
m
=
2
a
x
{\displaystyle m=2ax}
. Replacing x gives the parametric representation of the parabola with the tangent slope as parameter:
(
m
2
a
,
m
2
4
a
)
.
{\displaystyle \left({\tfrac {m}{2a}},{\tfrac {m^{2}}{4a}}\right)\!.}
The tangent has the equation
y
=
m
x
+
n
{\displaystyle y=mx+n}
with the still unknown n , which can be determined by inserting the coordinates of the parabola point. One gets
y
=
m
x
−
m
2
4
a
.
{\displaystyle y=mx-{\tfrac {m^{2}}{4a}}\;.}
If a tangent contains the point (x 0 , y 0 ) , off the parabola, then the equation
y
0
=
m
x
0
−
m
2
4
a
→
m
2
−
4
a
x
0
m
+
4
a
y
0
=
0
{\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
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
=
4
a
y
0
{\displaystyle m_{1}m_{2}=-1=4ay_{0}}
The last equation is equivalent to
y
0
=
−
1
4
a
,
{\displaystyle y_{0}=-{\frac {1}{4a}}\,,}
which is the equation of the
directrix .
Orthoptic of an ellipse and hyperbola [ edit ]
Ellipse [ edit ]
Let
E
:
x
2
a
2
+
y
2
b
2
=
1
{\displaystyle E:\;{\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1}
be the ellipse of consideration.
The tangents to the ellipse
E
{\displaystyle E}
at the vertices and co-vertices intersect at the 4 points
(
±
a
,
±
b
)
{\displaystyle (\pm a,\pm b)}
, which lie on the desired orthoptic curve (the circle
x
2
+
y
2
=
a
2
+
b
2
{\displaystyle x^{2}+y^{2}=a^{2}+b^{2}}
).
The tangent at a point
(
u
,
v
)
{\displaystyle (u,v)}
of the ellipse
E
{\displaystyle E}
has the equation
u
a
2
x
+
v
b
2
y
=
1
{\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 :
y
=
−
b
2
u
a
2
v
x
+
b
2
v
.
{\displaystyle y=-{\tfrac {b^{2}u}{a^{2}v}}\;x\;+\;{\tfrac {b^{2}}{v}}\,.}
Using the abbreviations
m
=
−
b
2
u
a
2
v
,
n
=
b
2
v
{\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
u
2
a
2
=
1
−
v
2
b
2
=
1
−
b
2
n
2
{\displaystyle {\color {blue}{\tfrac {u^{2}}{a^{2}}}=1-{\tfrac {v^{2}}{b^{2}}}=1-{\tfrac {b^{2}}{n^{2}}}}}
one gets:
m
2
=
b
4
u
2
a
4
v
2
=
1
a
2
b
4
v
2
u
2
a
2
=
1
a
2
n
2
(
1
−
b
2
n
2
)
=
n
2
−
b
2
a
2
.
{\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}}}\,.}
Hence
n
=
±
m
2
a
2
+
b
2
{\displaystyle n=\pm {\sqrt {m^{2}a^{2}+b^{2}}}}
(II )
and the equation of a non vertical tangent is
y
=
m
x
±
m
2
a
2
+
b
2
.
{\displaystyle y=mx\pm {\sqrt {m^{2}a^{2}+b^{2}}}.}
Solving relations
(I) for
u
,
v
{\displaystyle u,v}
and respecting
(II) leads to the slope depending parametric representation of the ellipse:
(
u
,
v
)
=
(
−
m
a
2
±
m
2
a
2
+
b
2
,
b
2
±
m
2
a
2
+
b
2
)
.
{\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)\,.}
(For another proof: see
Ellipse § Parametric representation .)
If a tangent contains the point
(
x
0
,
y
0
)
{\displaystyle (x_{0},y_{0})}
, off the ellipse, then the equation
y
0
=
m
x
0
±
m
2
a
2
+
b
2
{\displaystyle y_{0}=mx_{0}\pm {\sqrt {m^{2}a^{2}+b^{2}}}}
holds. Eliminating the square root leads to
m
2
−
2
x
0
y
0
x
0
2
−
a
2
m
+
y
0
2
−
b
2
x
0
2
−
a
2
=
0
,
{\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
m
1
,
m
2
{\displaystyle m_{1},m_{2}}
corresponding to the two tangents passing through
(
x
0
,
y
0
)
{\displaystyle (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
)
{\displaystyle (x_{0},y_{0})}
orthogonally, the following equations hold:
Orthoptics (red circles) of a circle, ellipses and hyperbolas
m
1
m
2
=
−
1
=
y
0
2
−
b
2
x
0
2
−
a
2
{\displaystyle 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
.
{\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
x
2
+
y
2
=
a
2
+
b
2
{\displaystyle x^{2}+y^{2}=a^{2}+b^{2}}
.
Hyperbola [ edit ]
The ellipse case can be adopted nearly exactly to the hyperbola case. The only changes to be made are to replace
b
2
{\displaystyle b^{2}}
with
−
b
2
{\displaystyle -b^{2}}
and to restrict m to |m | > b / a . Therefore:
The intersection points of orthogonal tangents are points of the circle
x
2
+
y
2
=
a
2
−
b
2
{\displaystyle x^{2}+y^{2}=a^{2}-b^{2}}
, where
a > b .
Orthoptic of an astroid [ edit ]
Orthoptic (purple) of an astroid
An astroid can be described by the parametric representation
c
(
t
)
=
(
cos
3
t
,
sin
3
t
)
,
0
≤
t
<
2
π
.
{\displaystyle \mathbf {c} (t)=\left(\cos ^{3}t,\sin ^{3}t\right),\quad 0\leq t<2\pi .}
From the condition
c
˙
(
t
)
⋅
c
˙
(
t
+
α
)
=
0
{\displaystyle \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:
y
=
−
tan
t
(
x
−
cos
3
t
)
+
sin
3
t
,
y
=
1
tan
t
(
x
+
sin
3
t
)
+
cos
3
t
.
{\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:
x
=
sin
t
cos
t
(
sin
t
−
cos
t
)
,
y
=
sin
t
cos
t
(
sin
t
+
cos
t
)
.
{\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}}}
This is simultaneously a parametric representation of the orthoptic.
Elimination of the parameter t yields the implicit representation
2
(
x
2
+
y
2
)
3
−
(
x
2
−
y
2
)
2
=
0.
{\displaystyle 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
x
=
1
2
cos
(
2
φ
)
cos
φ
,
y
=
1
2
cos
(
2
φ
)
sin
φ
.
{\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
r
=
1
2
cos
(
2
φ
)
,
0
≤
φ
<
2
π
{\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 [ edit ]
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 [ edit ]
Parabola:
The α -isoptics of the parabola with equation y = ax 2 are the branches of the hyperbola
x
2
−
tan
2
α
(
y
+
1
4
a
)
2
−
y
a
=
0.
{\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 x 2 / a 2 + y 2 / b 2 = 1 are the two parts of the degree-4 curve
(
x
2
+
y
2
−
a
2
−
b
2
)
2
tan
2
α
=
4
(
a
2
y
2
+
b
2
x
2
−
a
2
b
2
)
{\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 x 2 / a 2 − y 2 / b 2 = 1 are the two parts of the degree-4 curve
(
x
2
+
y
2
−
a
2
+
b
2
)
2
tan
2
α
=
4
(
a
2
y
2
−
b
2
x
2
+
a
2
b
2
)
.
{\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).}
Parabola:
A parabola y = ax 2 can be parametrized by the slope of its tangents m = 2ax :
c
(
m
)
=
(
m
2
a
,
m
2
4
a
)
,
m
∈
R
.
{\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
y
=
m
x
−
m
2
4
a
.
{\displaystyle y=mx-{\frac {m^{2}}{4a}}.}
The point (x 0 , y 0 ) is on the tangent if and only if
y
0
=
m
x
0
−
m
2
4
a
.
{\displaystyle 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
−
4
a
x
0
m
+
4
a
y
0
=
0.
{\displaystyle m^{2}-4ax_{0}m+4ay_{0}=0.}
If the tangents meet at angle α or 180° − α , the equation
tan
2
α
=
(
m
1
−
m
2
1
+
m
1
m
2
)
2
{\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 m 1 , m 2 into the last equation, one gets
x
0
2
−
tan
2
α
(
y
0
+
1
4
a
)
2
−
y
0
a
=
0.
{\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 x 2 / a 2 + y 2 / b 2 = 1 one can adopt the idea for the orthoptic for the quadratic equation
m
2
−
2
x
0
y
0
x
0
2
−
a
2
m
+
y
0
2
−
b
2
x
0
2
−
a
2
=
0.
{\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 m 1 , m 2 must be inserted into the equation
tan
2
α
=
(
m
1
−
m
2
1
+
m
1
m
2
)
2
.
{\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:
(
x
0
2
+
y
0
2
−
a
2
−
b
2
)
2
tan
2
α
=
4
(
a
2
y
0
2
+
b
2
x
0
2
−
a
2
b
2
)
.
{\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 b 2 with −b 2 (as in the case of the orthoptics, see above ).
To visualize the isoptics, see implicit curve .
External links [ edit ]
Wikimedia Commons has media related to
Isoptics .
References [ edit ]