Editing Hyperbola (section)

====Vertices====
In general the vectors <math>\vec f_1, \vec f_2</math> are not perpendicular. That means, in general <math>\vec f_0\pm \vec f_1</math> are ''not'' the vertices of the hyperbola. But <math>\vec f_1\pm \vec f_2</math> point into the directions of the asymptotes. The tangent vector at point <math>\vec p(t)</math> is
<math display="block">\vec p'(t) = \vec f_1\sinh t + \vec f_2\cosh t \ .</math>
Because at a vertex the tangent is perpendicular to the major axis of the hyperbola one gets the parameter <math>t_0</math> of a vertex from the equation
<math display="block">\vec p'(t)\cdot \left(\vec p(t) -\vec f_0\right) = \left(\vec f_1\sinh t + \vec f_2\cosh t\right) \cdot \left(\vec f_1 \cosh t +\vec f_2 \sinh t\right) = 0</math>
and hence from
<math display="block">\coth (2t_0)= -\tfrac{\vec f_1^{\, 2}+\vec f_2^{\, 2}}{2\vec f_1 \cdot \vec f_2} \ ,</math>
which yields

<math display="block">t_0=\tfrac{1}{4}\ln\tfrac{\left(\vec f_1-\vec f_2\right)^2}{\left(\vec f_1+\vec f_2\right)^2}.</math>

The formulae {{nowrap|<math>\cosh^2 x + \sinh^2 x = \cosh 2x</math>,}} {{nowrap|<math>2\sinh x \cosh x = \sinh 2x</math>,}} and <math>\operatorname{arcoth} x = \tfrac{1}{2}\ln\tfrac{x+1}{x-1}</math> were used.

The two ''vertices'' of the hyperbola are <math>\vec f_0\pm\left(\vec f_1\cosh t_0 +\vec f_2 \sinh t_0\right).</math>