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 ]