information:filtering, structural:reformatting
Theorem of Apollonios on conjugate diameters

For an ellipse with semi-axes a, b {\displaystyle a,\,b} the following is true:

Let the ellipse be in the canonical form with parametric equation p → (t) = (a cos t, b sin t). {\displaystyle {\vec {p}}(t)=(a\cos t,\,b\sin t).}

The two points c → 1 = p → (t), c → 2 = p → (t + π 2) {\textstyle {\vec {c}}_{1}={\vec {p}}(t),\ {\vec {c}}_{2}={\vec {p}}\left(t+{\frac {\pi }{2}}\right)} are on conjugate diameters (see previous section). From trigonometric formulae one obtains c → 2 = (− a sin t, b cos t) T {\displaystyle {\vec {c}}_{2}=(-a\sin t,\,b\cos t)^{\mathsf {T}}} and | c → 1 | 2 + | c → 2 | 2 = ⋯ = a 2 + b 2. {\displaystyle \left|{\vec {c}}_{1}\right|^{2}+\left|{\vec {c}}_{2}\right|^{2}=\cdots =a^{2}+b^{2}\,.}

The area of the triangle generated by c → 1, c → 2 {\displaystyle {\vec {c}}_{1},\,{\vec {c}}_{2}} is A Δ = 1 2 det (c → 1, c → 2) = ⋯ = 1 2 a b {\displaystyle A_{\Delta }={\tfrac {1}{2}}\det \left({\vec {c}}_{1},\,{\vec {c}}_{2}\right)=\cdots ={\tfrac {1}{2}}ab}

and from the diagram it can be seen that the area of the parallelogram is 8 times that of A Δ {\displaystyle A_{\Delta }}. Hence Area 12 = 4 a b. {\displaystyle {\text{Area}}_{12}=4ab\,.}