Dandelin's Theorem/Foci/Proof

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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:

$\SS$ and $\SS'$ are tangent to $\PP$ at the foci of $\EE$.


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 $\EE$.

$\Box$


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 $\EE$.

$\Box$


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.

$\blacksquare$


Theorem