Orthoptic (geometry)

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

1. The tangents to the ellipse ${\displaystyle E}$ at the vertices and co-vertices intersect at the 4 points ${\displaystyle (\pm a,\pm b)}$, which lie on the desired orthoptic curve (the circle ${\displaystyle x^{2}+y^{2}=a^{2}+b^{2}}$).
2. The tangent at a point ${\displaystyle (u,v)}$ of the ellipse ${\displaystyle E}$ has the equation ${\displaystyle {\tfrac {u}{a^{2}}}x+{\tfrac {v}{b^{2}}}y=1}$ (see tangent to an ellipse). If the point is not a vertex this equation can be solved for y: ${\displaystyle y=-{\tfrac {b^{2}u}{a^{2}v}}\;x\;+\;{\tfrac {b^{2}}{v}}\,.}$

Using the abbreviations

$${\displaystyle {\begin{aligned}m&=-{\tfrac {b^{2}u}{a^{2}v}},\\\color {red}n&=\color {red}{\tfrac {b^{2}}{v}}\end{aligned}}}$$

(I)

and the equation ${\displaystyle {\color {blue}{\tfrac {u^{2}}{a^{2}}}=1-{\tfrac {v^{2}}{b^{2}}}=1-{\tfrac {b^{2}}{n^{2}}}}}$ one gets:

$${\displaystyle m^{2}={\frac {b^{4}u^{2}}{a^{4}v^{2}}}={\frac {1}{a^{2}}}{\color {red}{\frac {b^{4}}{v^{2}}}}{\color {blue}{\frac {u^{2}}{a^{2}}}}={\frac {1}{a^{2}}}{\color {red}n^{2}}{\color {blue}\left(1-{\frac {b^{2}}{n^{2}}}\right)}={\frac {n^{2}-b^{2}}{a^{2}}}\,.}$$

Hence

$${\displaystyle n=\pm {\sqrt {m^{2}a^{2}+b^{2}}}}$$

(II)

and the equation of a non vertical tangent is

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

Solving relations (I) for ${\displaystyle u,v}$ and respecting (II) leads to the slope depending parametric representation of the ellipse:

$${\displaystyle (u,v)=\left(-{\tfrac {ma^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\;,\;{\tfrac {b^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\right)\,.}$$

(For another proof: see Ellipse § Parametric representation.)

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. Eliminating the square root leads to

$${\displaystyle m^{2}-{\frac {2x_{0}y_{0}}{x_{0}^{2}-a^{2}}}m+{\frac {y_{0}^{2}-b^{2}}{x_{0}^{2}-a^{2}}}=0,}$$

which has two solutions ${\displaystyle m_{1},m_{2}}$ corresponding to the two tangents passing through ${\displaystyle (x_{0},y_{0})}$. The constant term of a monic quadratic equation is always the product of its solutions. Hence, if the tangents meet at ${\displaystyle (x_{0},y_{0})}$ orthogonally, the following equations hold:

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

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

The last equation is equivalent to

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

From (1) and (2) one gets:

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

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

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

Parabola:

The α-isoptics of the parabola with equation y = ax² are the branches of the hyperbola

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

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

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

(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

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

Parabola:

A parabola y = ax² can be parametrized by the slope of its tangents m = 2ax:

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

The tangent with slope m has the equation

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

The point (x₀, y₀) is on the tangent if and only if

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

This means the slopes m₁, m₂ of the two tangents containing (x₀, y₀) fulfil the quadratic equation

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

If the tangents meet at angle α or 180° − α, the equation

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

must be fulfilled. Solving the quadratic equation for m, and inserting m₁, m₂ into the last equation, one gets

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

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

$${\displaystyle m^{2}-{\frac {2x_{0}y_{0}}{x_{0}^{2}-a^{2}}}m+{\frac {y_{0}^{2}-b^{2}}{x_{0}^{2}-a^{2}}}=0.}$$

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

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

Rearranging shows that the isoptics are parts of the degree-4 curve:

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

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.