Editing Ellipse (section)

=== 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>