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.


 * Dandelins-theorem-ellipse.png

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 $\EE$.

Proof for Hyperbola
Let $\EE$ be a hyperbola.


 * Dandelins-theorem-hyperbola.png

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 $\EE$.

Proof for Parabola
Let $\EE$ be a parabola.


 * Dandelins-theorem-parabola.png

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.

Theorem

 * : Chapter $\text {IV}$. The Ellipse: $1 \text a$. Focal properties