Construction of Conic Section

Theorem
Consider a right circular cone $\CC$ with opening angle $2 \alpha$ whose apex is at $O$.

Consider a plane $\PP$ which intersects $\CC$, not passing through $O$, at an angle $\beta$ to the axis of $\CC$.

Then the points of intersection form one of the following:
 * a circle
 * an ellipse
 * a hyperbola
 * a parabola.

Proof
When $\beta$ is a right angle, the points of intersection form a circle, by definition of transverse section.

Otherwise, let the plane $OAA'$ through the axis of $\CC$ perpendicular to $\PP$ intersect $\PP$ in the line $AA'$.

Let $P$ be an arbitrary point on the intersection of $\PP$ with $\CC$.

Let $PM$ be constructed perpendicular to $AA'$.

$PM$ is then perpendicular to the plane $OAA'$.

The transverse section through $P$ then contains $PM$ and cuts $OAA'$ in $TT'$, which is a diameter of that transverse section.

The diagram below is the cross-section through $\CC$ corresponding to the plane $OAA'$.

The point $P$ is imagined to be perpendicular to the plane whose projection onto $OAA'$ coincides with $M$.


 * Conic-section-construction.png

Because the transverse section is a circle, we have by the Intersecting Chord Theorem:
 * $PM^2 = TM \cdot MT'$

Also:
 * $\dfrac {TM} {AM} = \dfrac {\map \sin {\beta + \alpha} } {\cos \alpha}$
 * $\dfrac {MT'} {MA'} = \dfrac {\map \sin {\beta - \alpha} } {\cos \alpha}$

and so:
 * $PM^2 = k \cdot AM \cdot MA'$

where $k$ is the constant:
 * $k = \dfrac {\map \sin {\beta + \alpha} \map \sin {\beta - \alpha} } {\cos \alpha}$

When $\beta > \alpha$, $\PP$ intersects one nappe of $\CC$.

In this case the intersection is a closed curve which is the ellipse, and $k$ is positive.

The plane passing through $O$ which is parallel to $\PP$ intersects $\CC$ at $O$ only.

When $\beta < \alpha$, $\PP$ intersects both nappes of $\CC$.

In this case the intersection consists of two open curves which together form the hyperbola, and $k$ is negative.

The plane passing through $O$ which is parallel to $\PP$ intersects $\CC$ in two straight lines which themselves intersect at $O$.

When $\beta = \alpha$, $\PP$ intersects one nappes of $\CC$.

In this case the intersection consists of one open curves which is the parabola, and $k$ is zero.

The plane passing through $O$ which is parallel to $\PP$ intersects $\CC$ in one straight line, which is a generatrix of $\CC$.

Hence:
 * $MT' = 2 OA \sin \alpha$

which is constant.

Also:
 * $TM = 2 AM \sin \alpha$

and so:
 * $PM^2 = 4 OA \sin^2 \alpha \cdot AM$

That is:
 * $PM^2 : AM$ is constant.