Any parabola can be transformed by a rigid motion (angles are not changed) into a parabola with equation . The slope at a point of the parabola is . Replacing x gives the parametric representation of the parabola with the tangent slope as parameter: The tangent has the equation with the still unknown n, which can be determined by inserting the coordinates of the parabola point. One gets
If a tangent contains the point (x0, y0), off the parabola, then the equation
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:
The last equation is equivalent to
which is the equation of the directrix.
The tangents to the ellipse at the vertices and co-vertices intersect at the 4 points , which lie on the desired orthoptic curve (the circle ).
The tangent at a point of the ellipse has the equation (see tangent to an ellipse). If the point is not a vertex this equation can be solved for y:
Using the abbreviations
(I)
and the equation one gets:
Hence
(II)
and the equation of a non vertical tangent is
Solving relations (I) for and respecting (II) leads to the slope depending parametric representation of the ellipse:
(For another proof: see Ellipse § Parametric representation.)
If a tangent contains the point , off the ellipse, then the equation
holds. Eliminating the square root leads to
which has two solutions corresponding to the two tangents passing through . The constant term of a monic quadratic equation is always the product of its solutions. Hence, if the tangents meet at orthogonally, the following equations hold:
The last equation is equivalent to
From (1) and (2) one gets:
The intersection points of orthogonal tangents are points of the circle .
The ellipse case can be adopted nearly exactly to the hyperbola case. The only changes to be made are to replace with and to restrict m to |m| > b/a. Therefore:
The intersection points of orthogonal tangents are points of the circle , where a > b.
An astroid can be described by the parametric representation
From the condition
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:
Their common point has coordinates:
This is simultaneously a parametric representation of the orthoptic.
Elimination of the parameter t yields the implicit representation
Introducing the new parameter φ = t − 5π/4 one gets
(The proof uses the angle sum and difference identities.) Hence we get the polar representation
of the orthoptic. Hence:
The α-isoptics of the parabola with equation y = ax2 are the branches of the hyperbola
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
(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