Dandelin's Theorem/Foci/Proof
Theorem
Let $\CC$ be a double napped right circular cone with apex $O$.
Let $\PP$ be a plane which intersects $\CC$ such that:
- $\PP$ does not pass through $O$
- $\PP$ is not perpendicular to the axis of $\CC$.
Let $\EE$ be the conic section arising as the intersection between $\PP$ and $\CC$.
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Then:
Proof
Let $\SS$ and $\SS'$ be the Dandelin spheres with respect to $\PP$.
Let $P$ be a point on $\EE$.
Let $F$ and $F'$ be the points at which $\SS$ and $\SS'$ are tangent to $\PP$ respectively.
Let the generatrix of $\CC$ which passes through $P$ touch $\SS$ and $\SS'$ at $E$ and $E'$ respectively.
Because $PF$ and $PE$ are both tangent to $\SS$:
- $PF = PE$
and similarly:
- $PF' = PE'$
Proof for Ellipse
Let $\EE$ be an ellipse.
Then $E$ and $E'$ are on the same side of $O$, and so:
- $PF + PF' = PE + PE'$
But $PE + PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
$\Box$
Proof for Hyperbola
Let $\EE$ be a hyperbola.
Then $E$ and $E'$ are on the opposites sides of $O$, and so:
- $PF - PF' = PE - PE'$
But $PE - PE'$ is a constant.
Hence $\EE$ fulfils the equidistance property with respect to the points $F$ and $F'$.
Hence $F$ and $F'$ are the foci of $\EE$.
$\Box$
Proof for Parabola
Let $\EE$ be a parabola.
Because $\PP$ is parallel to the generatrix of $\CC$, it follows that:
- $\triangle PHE = \triangle PHE'$
and so:
- $PF = PE'$
and it is seen that this is the focus-directrix property of the parabola.
$\blacksquare$
Theorem
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {IV}$. The Ellipse: $1 \text a$. Focal properties