Hyperbola
Tangent–asymptotes triangle

For simplicity the center of the hyperbola may be the origin and the vectors f → 1, f → 2 {\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}} have equal length. If the last assumption is not fulfilled one can first apply a parameter transformation (see above) in order to make the assumption true. Hence ± (f → 1 + f → 2) {\displaystyle \pm ({\vec {f}}_{1}+{\vec {f}}_{2})} are the vertices, ± (f → 1 − f → 2) {\displaystyle \pm ({\vec {f}}_{1}-{\vec {f}}_{2})} span the minor axis and one gets | f → 1 + f → 2 | = a {\displaystyle |{\vec {f}}_{1}+{\vec {f}}_{2}|=a} and | f → 1 − f → 2 | = b {\displaystyle |{\vec {f}}_{1}-{\vec {f}}_{2}|=b}.

For the intersection points of the tangent at point p → (t 0) = f → 1 t 0 + f → 2 1 t 0 {\displaystyle {\vec {p}}(t_{0})={\vec {f}}_{1}t_{0}+{\vec {f}}_{2}{\tfrac {1}{t_{0}}}} with the asymptotes one gets the points C = 2 t 0 f → 1, D = 2 t 0 f → 2. {\displaystyle C=2t_{0}{\vec {f}}_{1},\ D={\tfrac {2}{t_{0}}}{\vec {f}}_{2}.} The area of the triangle M, C, D {\displaystyle M,C,D} can be calculated by a 2 × 2 determinant: A = 1 2 | det (2 t 0 f → 1, 2 t 0 f → 2) | = 2 | det (f → 1, f → 2) | {\displaystyle A={\tfrac {1}{2}}{\Big |}\det \left(2t_{0}{\vec {f}}_{1},{\tfrac {2}{t_{0}}}{\vec {f}}_{2}\right){\Big |}=2{\Big |}\det \left({\vec {f}}_{1},{\vec {f}}_{2}\right){\Big |}} (see rules for determinants). | det (f → 1, f → 2) | {\displaystyle \left|\det({\vec {f}}_{1},{\vec {f}}_{2})\right|} is the area of the rhombus generated by f → 1, f → 2 {\displaystyle {\vec {f}}_{1},{\vec {f}}_{2}}. The area of a rhombus is equal to one half of the product of its diagonals. The diagonals are the semi-axes a, b {\displaystyle a,b} of the hyperbola. Hence:
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