Focal conics

Ellipse and hyperbola ;Equations If one describes the ellipse in the x-y-plane in the common way by the equation :\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\; , then the corresponding focal hyperbola in the x-z-plane has equation :\frac{x^2}{c^2} - \frac{z^2}{b^2} = 1\; , where c is the linear eccentricity of the ellipse with \; c^2 = a^2 - b^2\; . ;Parametric representations :ellipse: \quad \vec u(\varphi)=(a\cos \varphi,b\sin \varphi,0)^T\ and :hyperbola: \ \vec v(\psi)=(c\cosh \psi, 0,b\sinh \psi)^T\ . Two parabolas Two parabolas in the x-y-plane and in the x-z-plane: :1. parabola: \ y^2=p^2-2px\ and :2. parabola: \ z^2=2px \ . with p the semi-latus rectum of both the parabolas. == Right circular cones through an ellipse ==

• The apices of the right circular cones through a given ellipse lie on the focal hyperbola belonging to the ellipse. ;Proof Given: Ellipse with vertices A,B and foci E,F and a right circular cone with apex S containing the ellipse (see diagram). Because of symmetry the axis of the cone has to be contained in the plane through the foci, which is orthogonal to the ellipse's plane. There exists a Dandelin sphere k, which touches the ellipse's plane at the focus F and the cone at a circle. From the diagram and the fact that all tangential distances of a point to a sphere are equal one gets: :|AS|=|AA_1|+|A_1S|=|AF|+|B_1S| :|BS|=|BB_1|+|B_1S|=|BF|+|B_1S| Hence: :|AS|-|BS|=|AF|-|BF|=|EF|= const. and the set of all possible apices lie on the hyperbola with the vertices E,F and the foci A,B. Analogously one proves the cases, where the cones contain a hyperbola or a parabola. == References ==