Orthoptic (geometry)

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.

Examples:

1. The orthoptic of a parabola is its directrix (proof: see below),
2. The orthoptic of an ellipse ${\displaystyle {\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1}${\displaystyle {\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1} is the director circle ${\displaystyle x^{2}+y^{2}=a^{2}+b^{2}}${\displaystyle x^{2}+y^{2}=a^{2}+b^{2}} (see below),
3. The orthoptic of a hyperbola ${\displaystyle {\tfrac {x^{2}}{a^{2}}}-{\tfrac {y^{2}}{b^{2}}}=1,\ a>b}${\displaystyle {\tfrac {x^{2}}{a^{2}}}-{\tfrac {y^{2}}{b^{2}}}=1,\ a>b} is the director circle ${\displaystyle x^{2}+y^{2}=a^{2}-b^{2}}${\displaystyle x^{2}+y^{2}=a^{2}-b^{2}} (in case of a ≤ b there are no orthogonal tangents, see below),
4. The orthoptic of an astroid ${\displaystyle x^{2/3}+y^{2/3}=1}${\displaystyle x^{2/3}+y^{2/3}=1} is a quadrifolium with the polar equation ${\displaystyle r={\tfrac {1}{\sqrt {2}}}\cos(2\varphi ),\ 0\leq \varphi <2\pi }${\displaystyle r={\tfrac {1}{\sqrt {2}}}\cos(2\varphi ),\ 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.

Any parabola can be transformed by a rigid motion (angles are not changed) into a parabola with equation ${\displaystyle y=ax^{2}}${\displaystyle y=ax^{2}}. The slope at a point of the parabola is ${\displaystyle m=2ax}${\displaystyle m=2ax}. Replacing x gives the parametric representation of the parabola with the tangent slope as parameter: ${\displaystyle \left({\tfrac {m}{2a}},{\tfrac {m^{2}}{4a}}\right)\!.}${\displaystyle \left({\tfrac {m}{2a}},{\tfrac {m^{2}}{4a}}\right)\!.} The tangent has the equation ${\displaystyle y=mx+n}${\displaystyle y=mx+n} with the still unknown n, which can be determined by inserting the coordinates of the parabola point. One gets ${\displaystyle y=mx-{\tfrac {m^{2}}{4a}}\;.}${\displaystyle y=mx-{\tfrac {m^{2}}{4a}}\;.}

If a tangent contains the point (x₀, y₀), off the parabola, then the equation $${\displaystyle y_{0}=mx_{0}-{\frac {m^{2}}{4a}}\quad \rightarrow \quad m^{2}-4ax_{0},円m+4ay_{0}=0}$${\displaystyle y_{0}=mx_{0}-{\frac {m^{2}}{4a}}\quad \rightarrow \quad m^{2}-4ax_{0},円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: $${\displaystyle m_{1}m_{2}=-1=4ay_{0}}$${\displaystyle m_{1}m_{2}=-1=4ay_{0}} The last equation is equivalent to $${\displaystyle y_{0}=-{\frac {1}{4a}},,円}$${\displaystyle y_{0}=-{\frac {1}{4a}},,円} which is the equation of the directrix.

Let ${\displaystyle E:\;{\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1}${\displaystyle E:\;{\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1} be the ellipse of consideration.

1. The tangents to the ellipse ${\displaystyle E}${\displaystyle E} at the vertices and co-vertices intersect at the 4 points ${\displaystyle (\pm a,\pm b)}${\displaystyle (\pm a,\pm b)}, which lie on the desired orthoptic curve (the circle ${\displaystyle x^{2}+y^{2}=a^{2}+b^{2}}${\displaystyle x^{2}+y^{2}=a^{2}+b^{2}}).
2. The tangent at a point ${\displaystyle (u,v)}${\displaystyle (u,v)} of the ellipse ${\displaystyle E}${\displaystyle E} has the equation ${\displaystyle {\tfrac {u}{a^{2}}}x+{\tfrac {v}{b^{2}}}y=1}${\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: ${\displaystyle y=-{\tfrac {b^{2}u}{a^{2}v}}\;x\;+\;{\tfrac {b^{2}}{v}},円.}${\displaystyle y=-{\tfrac {b^{2}u}{a^{2}v}}\;x\;+\;{\tfrac {b^{2}}{v}},円.}

Using the abbreviations

and the equation ${\displaystyle {\color {blue}{\tfrac {u^{2}}{a^{2}}}=1-{\tfrac {v^{2}}{b^{2}}}=1-{\tfrac {b^{2}}{n^{2}}}}}${\displaystyle {\color {blue}{\tfrac {u^{2}}{a^{2}}}=1-{\tfrac {v^{2}}{b^{2}}}=1-{\tfrac {b^{2}}{n^{2}}}}} one gets: $${\displaystyle m^{2}={\frac {b^{4}u^{2}}{a^{4}v^{2}}}={\frac {1}{a^{2}}}{\color {red}{\frac {b^{4}}{v^{2}}}}{\color {blue}{\frac {u^{2}}{a^{2}}}}={\frac {1}{a^{2}}}{\color {red}n^{2}}{\color {blue}\left(1-{\frac {b^{2}}{n^{2}}}\right)}={\frac {n^{2}-b^{2}}{a^{2}}},円.}$${\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}}},円.} Hence

and the equation of a non vertical tangent is $${\displaystyle y=mx\pm {\sqrt {m^{2}a^{2}+b^{2}}}.}$${\displaystyle y=mx\pm {\sqrt {m^{2}a^{2}+b^{2}}}.} Solving relations (I) for ${\displaystyle u,v}${\displaystyle u,v} and respecting (II) leads to the slope depending parametric representation of the ellipse: $${\displaystyle (u,v)=\left(-{\tfrac {ma^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\;,\;{\tfrac {b^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\right),円.}$${\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),円.} (For another proof: see Ellipse § Parametric representation.)

If a tangent contains the point ${\displaystyle (x_{0},y_{0})}${\displaystyle (x_{0},y_{0})}, off the ellipse, then the equation $${\displaystyle y_{0}=mx_{0}\pm {\sqrt {m^{2}a^{2}+b^{2}}}}$${\displaystyle y_{0}=mx_{0}\pm {\sqrt {m^{2}a^{2}+b^{2}}}} holds. Eliminating the square root leads to $${\displaystyle m^{2}-{\frac {2x_{0}y_{0}}{x_{0}^{2}-a^{2}}}m+{\frac {y_{0}^{2}-b^{2}}{x_{0}^{2}-a^{2}}}=0,}$${\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 ${\displaystyle m_{1},m_{2}}${\displaystyle m_{1},m_{2}} corresponding to the two tangents passing through ${\displaystyle (x_{0},y_{0})}${\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 ${\displaystyle (x_{0},y_{0})}${\displaystyle (x_{0},y_{0})} orthogonally, the following equations hold:

$${\displaystyle m_{1}m_{2}=-1={\frac {y_{0}^{2}-b^{2}}{x_{0}^{2}-a^{2}}}}$${\displaystyle m_{1}m_{2}=-1={\frac {y_{0}^{2}-b^{2}}{x_{0}^{2}-a^{2}}}} The last equation is equivalent to $${\displaystyle x_{0}^{2}+y_{0}^{2}=a^{2}+b^{2},円.}$${\displaystyle x_{0}^{2}+y_{0}^{2}=a^{2}+b^{2},円.} From (1) and (2) one gets:

The ellipse case can be adopted nearly exactly to the hyperbola case. The only changes to be made are to replace ${\displaystyle b^{2}}${\displaystyle b^{2}} with ${\displaystyle -b^{2}}${\displaystyle -b^{2}} and to restrict m to |m| > ⁠b/a⁠. Therefore:

An astroid can be described by the parametric representation $${\displaystyle \mathbf {c} (t)=\left(\cos ^{3}t,\sin ^{3}t\right),\quad 0\leq t<2\pi .}$${\displaystyle \mathbf {c} (t)=\left(\cos ^{3}t,\sin ^{3}t\right),\quad 0\leq t<2\pi .} From the condition $${\displaystyle \mathbf {\dot {c}} (t)\cdot \mathbf {\dot {c}} (t+\alpha )=0}$${\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: $${\displaystyle {\begin{aligned}y&=-\tan t\left(x-\cos ^{3}t\right)+\sin ^{3}t,\\y&={\frac {1}{\tan t}}\left(x+\sin ^{3}t\right)+\cos ^{3}t.\end{aligned}}}$${\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: $${\displaystyle {\begin{aligned}x&=\sin t\cos t\left(\sin t-\cos t\right),\\y&=\sin t\cos t\left(\sin t+\cos t\right).\end{aligned}}}$${\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 $${\displaystyle 2\left(x^{2}+y^{2}\right)^{3}-\left(x^{2}-y^{2}\right)^{2}=0.}$${\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 $${\displaystyle {\begin{aligned}x&={\tfrac {1}{\sqrt {2}}}\cos(2\varphi )\cos \varphi ,\\y&={\tfrac {1}{\sqrt {2}}}\cos(2\varphi )\sin \varphi .\end{aligned}}}$${\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 $${\displaystyle r={\tfrac {1}{\sqrt {2}}}\cos(2\varphi ),\quad 0\leq \varphi <2\pi }$${\displaystyle r={\tfrac {1}{\sqrt {2}}}\cos(2\varphi ),\quad 0\leq \varphi <2\pi } of the orthoptic. Hence:

Below the isotopics for angles α ≠ 90° are listed. They are called α-isoptics. For the proofs see below.

Parabola:

The α-isoptics of the parabola with equation y = ax² are the branches of the hyperbola $${\displaystyle x^{2}-\tan ^{2}\alpha \left(y+{\frac {1}{4a}}\right)^{2}-{\frac {y}{a}}=0.}$${\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 $${\displaystyle \left(x^{2}+y^{2}-a^{2}-b^{2}\right)^{2}\tan ^{2}\alpha =4\left(a^{2}y^{2}+b^{2}x^{2}-a^{2}b^{2}\right)}$${\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/a²⁠ − ⁠y²/b²⁠ = 1 are the two parts of the degree-4 curve $${\displaystyle \left(x^{2}+y^{2}-a^{2}+b^{2}\right)^{2}\tan ^{2}\alpha =4\left(a^{2}y^{2}-b^{2}x^{2}+a^{2}b^{2}\right).}$${\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 = ax² can be parametrized by the slope of its tangents m = 2ax: $${\displaystyle \mathbf {c} (m)=\left({\frac {m}{2a}},{\frac {m^{2}}{4a}}\right),\quad m\in \mathbb {R} .}$${\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 $${\displaystyle y=mx-{\frac {m^{2}}{4a}}.}$${\displaystyle y=mx-{\frac {m^{2}}{4a}}.}

The point (x₀, y₀) is on the tangent if and only if $${\displaystyle y_{0}=mx_{0}-{\frac {m^{2}}{4a}}.}$${\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 $${\displaystyle m^{2}-4ax_{0}m+4ay_{0}=0.}$${\displaystyle m^{2}-4ax_{0}m+4ay_{0}=0.}

If the tangents meet at angle α or 180° − α, the equation $${\displaystyle \tan ^{2}\alpha =\left({\frac {m_{1}-m_{2}}{1+m_{1}m_{2}}}\right)^{2}}$${\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 $${\displaystyle x_{0}^{2}-\tan ^{2}\alpha \left(y_{0}+{\frac {1}{4a}}\right)^{2}-{\frac {y_{0}}{a}}=0.}$${\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 $${\displaystyle m^{2}-{\frac {2x_{0}y_{0}}{x_{0}^{2}-a^{2}}}m+{\frac {y_{0}^{2}-b^{2}}{x_{0}^{2}-a^{2}}}=0.}$${\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 $${\displaystyle \tan ^{2}\alpha =\left({\frac {m_{1}-m_{2}}{1+m_{1}m_{2}}}\right)^{2}.}$${\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: $${\displaystyle \left(x_{0}^{2}+y_{0}^{2}-a^{2}-b^{2}\right)^{2}\tan ^{2}\alpha =4\left(a^{2}y_{0}^{2}+b^{2}x_{0}^{2}-a^{2}b^{2}\right).}$${\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.

• Special Plane Curves.
• Mathworld
• Jan Wassenaar's Curves
• "Isoptic curve" at MathCurve
• "Orthoptic curve" at MathCurve

• Lawrence, J. Dennis (1972). A catalog of special plane curves . Dover Publications. pp. 58–59. ISBN 0-486-60288-5.
• Odehnal, Boris (2010). "Equioptic Curves of Conic Sections" (PDF). Journal for Geometry and Graphics. 14 (1): 29–43.
• Schaal, Hermann (1977). Lineare Algebra und Analytische Geometrie. Vol. III. Vieweg. p. 220. ISBN 3-528-03058-5.
• Steiner, Jacob (1867). Vorlesungen über synthetische Geometrie. Leipzig: B. G. Teubner. Part 2, p. 186.
• Ternullo, Maurizio (2009). "Two new sets of ellipse related concyclic points". Journal of Geometry. 94 (1–2): 159–173. doi:10.1007/s00022-009-0005-7. S2CID 120011519.

• Curves

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