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Orthoptic (geometry)

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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.

Contents

Examples:

  1. The orthoptic of a parabola is its directrix (proof: see parabola),
  2. The orthoptic of an ellipse x2/a2 + y2/b2 = 1 is the director circle x2 + y2 = a2 + b2 (see below),
  3. The orthoptic of a hyperbola x2/a2 − y2/b2 = 1, a > b, is the circle x2 + y2 = a2 − b2 (in case of a ≤ b there are no orthogonal tangents, see below),
  4. The orthoptic of an astroid x 2⁄3 + y 2⁄3 = 1 is a quadrifolium with the polar equation

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.

Ellipse

The ellipse with equation

x 2 a 2 + y 2 b 2 = 1

can be represented by the unusual parametric representation

c → ± ( m ) = ( − m a 2 ± m 2 a 2 + b 2 , b 2 ± m 2 a 2 + b 2 ) , m ∈ R ,

where m is the slope of the tangent at a point of the ellipse. c→+(m) describes the upper half and c→−(m) the lower half of the ellipse. The points (±a, 0)) with tangents parallel to the y-axis are excluded. But this is no problem, because these tangents meet orthogonal the tangents parallel to the x-axis in the ellipse points (0, ±b) Hence the points (±a, ±b) are points of the desired orthoptic (the circle x2 + y2 = a2 + b2).

The tangent at point c→±(m) has the equation

y = m x ± m 2 a 2 + b 2 .

If a tangent contains the point (x0, y0), off the ellipse, then the equation

y 0 = m x 0 ± 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 ,

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:

m 1 m 2 = − 1 = 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

This means that:

  • The intersection points of orthogonal tangents are points of the circle x2 + y2 = a2 + b2.

    • Hyperbola

    The ellipse case can be adopted nearly exactly to the hyperbola case. The only changes to be made are to replace b2 with −b2 and to restrict m to | m | > b/a. Therefore:

  • The intersection points of orthogonal tangents are points of the circle x2 + y2 = a2 − b2, where a > b.

    • Orthoptic of an astroid

    An astroid can be described by the parametric representation

    c → ( t ) = ( cos 3 ⁡ t , sin 3 ⁡ t ) , 0 ≤ t < 2 π .

    From the condition

    c ˙ → ( t ) ⋅ c ˙ → ( t + α ) = 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 .

    Their common point has coordinates:

    x = sin ⁡ t cos ⁡ t ( sin ⁡ t − cos ⁡ t ) , y = sin ⁡ t cos ⁡ t ( sin ⁡ t + cos ⁡ t ) .

    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.

    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 .

    (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

    of the orthoptic. Hence:

  • The orthoptic of an astroid is a quadrifolium.

    • Isoptic of a parabola, an ellipse and a hyperbola

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

    Equations of the isoptics

    Parabola:

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

    x 2 − tan 2 ⁡ α ( y + 1 4 a ) 2 − 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

    ( 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 )

    (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

    ( 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 ) .

    Proofs

    Parabola:

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

    c → ( m ) = ( m 2 a , m 2 4 a ) , m ∈ R .

    The tangent with slope m has the equation

    y = m x − m 2 4 a .

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

    y 0 = m x 0 − m 2 4 a .

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

    m 2 − 4 a x 0 m + 4 a y 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

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

    x 0 2 − tan 2 ⁡ α ( y 0 + 1 4 a ) 2 − 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

    m 2 − 2 x 0 y 0 x 0 2 − a 2 m + 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

    tan 2 ⁡ α = ( m 1 − m 2 1 + m 1 m 2 ) 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 ) .
    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.

    References

    Orthoptic (geometry) Wikipedia
    (Text) CC BY-SA


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