Dandelin's Theorem/Directrices/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.

Let $\theta$ be half the opening angle of $\CC$.

Let $\phi$ be the inclination of $\PP$ to the axis of $\CC$.

Let $\PP$ intersect $\KK$ in the straight line $NX$.

Let $PN$ be constructed perpendicular to $NX$.

Let $PK$ be constructed perpendicular to $\KK$.

Then:
 * $PK = PN \cos \phi$

Also:
 * $PK = PE \cos \theta = PF \cos \theta$

Hence:
 * $\dfrac {PF} {PN} = \dfrac {\cos \phi} {\cos \theta}$

which is constant.

Hence $NX$ is a directrix of $\EE$ by definition.

A similar argument and construction applies with respect to $\SS'$ and $\KK$.

Theorem

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