Editing Ellipse (section)

== Conjugate diameters ==
=== Definition of conjugate diameters ===
[[File:Parallelproj-kreis-ellipse.svg|400px|thumb|Orthogonal diameters of a circle with a square of tangents, midpoints of parallel chords and an affine image, which is an ellipse with conjugate diameters, a parallelogram of tangents and midpoints of chords.]]
{{Main|Conjugate diameters}}

A circle has the following property:
: The midpoints of parallel chords lie on a diameter.

An affine transformation preserves parallelism and midpoints of line segments, so this property is true for any ellipse. (Note that the parallel chords and the diameter are no longer orthogonal.)

; Definition:
Two diameters <math>d_1,\, d_2</math> of an ellipse are ''conjugate'' if the midpoints of chords parallel to <math>d_1</math> lie on <math>d_2\ .</math>

From the diagram one finds:
: Two diameters <math>\overline{P_1 Q_1},\, \overline{P_2 Q_2}</math> of an ellipse are conjugate whenever the tangents at <math>P_1</math> and <math>Q_1</math> are parallel to <math>\overline{P_2 Q_2}</math>.

Conjugate diameters in an ellipse generalize orthogonal diameters in a circle.

In the parametric equation for a general ellipse given above,
<math display="block">\vec x = \vec p(t) = \vec f\!_0 +\vec f\!_1 \cos t + \vec f\!_2 \sin t,</math>

any pair of points <math>\vec p(t),\ \vec p(t + \pi)</math> belong to a diameter, and the pair <math>\vec p\left(t + \tfrac{\pi}{2}\right),\ \vec p\left(t - \tfrac{\pi}{2}\right)</math> belong to its conjugate diameter.

For the common parametric representation <math>(a\cos t,b\sin t)</math> of the ellipse with equation <math>\tfrac{x^2}{a^2}+\tfrac{y^2}{b^2}=1</math> one gets: The points 
:<math>(x_1,y_1)=(\pm a\cos t,\pm b\sin t)\quad </math> (signs: (+,+) or (−,−) )
:<math>(x_2,y_2)=({\color{red}{\mp}} a\sin t,\pm b\cos t)\quad </math> (signs: (−,+) or (+,−) )
:are conjugate and 
:<math>\frac{x_1x_2}{a^2}+\frac{y_1y_2}{b^2}=0\ .</math>
In case of a circle the last equation collapses to <math>x_1x_2+y_1y_2=0\ . </math>

=== Theorem of Apollonios on conjugate diameters ===
[[File:Elli-apoll-cd.svg|upright=1.2|thumb|Theorem of Apollonios]]
[[File:Elli-apoll-area-altern.svg|thumb|upright=1.2|For the alternative area formula]]
For an ellipse with semi-axes <math>a,\, b</math> the following is true:<ref>Bronstein&Semendjajew: ''Taschenbuch der Mathematik'', Verlag Harri Deutsch, 1979, {{ISBN|3871444928}}, p. 274.</ref><ref>''Encyclopedia of Mathematics'', Springer,  URL: http://encyclopediaofmath.org/index.php?title=Apollonius_theorem&oldid=17516 .</ref>

: Let <math>c_1 </math> and  <math> c_2</math> be halves of two conjugate diameters (see diagram) then
:# <math>c_1^2 + c_2^2 = a^2 + b^2</math>.
:# The ''triangle'' <math>O,P_1,P_2</math> with sides <math>c_1,\, c_2</math> (see diagram) has the constant area <math display="inline">A_\Delta = \frac{1}{2}ab</math>, which can be expressed by <math>A_\Delta=\tfrac 1 2 c_2d_1=\tfrac 1 2 c_1c_2\sin\alpha</math>, too. <math>d_1</math> is the altitude of point <math>P_1</math> and <math>\alpha</math> the angle between the half diameters. Hence the area of the ellipse (see section [[#Metric properties|metric properties]]) can be written as <math>A_{el}=\pi ab=\pi c_2d_1=\pi c_1c_2\sin\alpha</math>.
:# The parallelogram of tangents adjacent to the given conjugate diameters has the <math>\text{Area}_{12} = 4ab\ .</math>

; Proof:
Let the ellipse be in the canonical form with parametric equation
<math display="block">\vec p(t) = (a\cos t,\, b\sin t).</math>

The two points <math display="inline">\vec c_1 = \vec p(t),\ \vec c_2 = \vec p\left(t + \frac{\pi}{2}\right)</math> are on conjugate diameters (see previous section). From trigonometric formulae one obtains <math>\vec c_2 = (-a\sin t,\, b\cos t)^\mathsf{T}</math> and
<math display="block">\left|\vec c_1\right|^2 + \left|\vec c_2\right|^2 = \cdots = a^2 + b^2\, .</math>

The area of the triangle generated by <math>\vec c_1,\, \vec c_2</math> is
<math display="block">A_\Delta = \tfrac{1}{2} \det\left(\vec c_1,\, \vec c_2\right) = \cdots = \tfrac{1}{2}ab</math>

and from the diagram it can be seen that the area of the parallelogram is 8 times that of <math>A_\Delta</math>. Hence
<math display="block">\text{Area}_{12} = 4ab\, .</math>