Is a parabola really a conic section?

Question

A parabola is supposed to be a conic section, obtained by slicing through a cone with a plane, like this from Wikipedia:

But if you do this, what you get is just part of an elipse, since if the cone and the plane were bigger, eventually you'd end up with an elipse. And a section of an elipse is not a parabola - there is no possible focal point that satisfies the condition that all points on the curve are equidistant from the focal point and the directrix - at least not when I try to construct this in my CAD software

Answered by JonathanZ, the key thing was the parabola needs the slicing plane parallel to the slope of the cone (not illustrated or explained clearly in either Wikipedia or Wolfram IMHO). Here's what it looks like in my construction:

Answer 1

It's not clear what you're saying about "not being able to find a focal point and directrix", but if you are trying to create this construction with CAD software, it seems worth mentioning that "being a parabola" exists at a "critical point", i.e.

• if you've got an ellipse and jiggle parameters a small bit, you still get an ellipse,

• if you've got a hyperbola and jiggle parameters a small bit, you still get a hyperbola

• if you've got a parabola and jiggle parameters a small bit, jiggle one direction and you get an ellipse, jiggle in the other direction and you get a hyperbola.

Phenomena like this, which are perched on the transition line between two areas, are very hard to render with finite precision math. I wonder if that's the problem you're running into?

Also, as to your comment "if the cone and the plane were bigger, eventually you'd end up with an elipse" - in order to get a parabola, the slicing plane has to be exactly "parallel" to the edge of the cone opposite to it, i.e. never intersect it. (It looks a little off in your image, possibly due to the angle of view.) This is again something very hard to do with finite precision math: Draw some random line, then try to come up with a plane that never intersects it. If you're the least bit off there's going to be some intersection, probably way off in the distance, but intersection none the less - even though we can mathematically show such a plane exists, and give an equation for it.

Answer 2

If the plane is parallel to the slope of the cone, then no matter how big the plane and the cone are, it will never close into an elipse.

Take the vertex of the parabola in your image, now draw a line perpendicular to the parabola (i.e., along the axis of symmetry, in the image it is the dotted straight line). If the intersection was an elipse, this line should eventually intersect again the cone. But since it is parallel to the slope of the cone, it will never cut it again.

Answer 3

Let us work out the equation of the intersection of a cone and plane in terms of some plane coordinates $(u,v)$ and see if it is a parabola indeed. Place a coordinate system on the apex of the cone.

Place the axis of the cone along the $\hat{j}$ direction, and slice it with a plane rotated an angle half of the cone angle $\psi$

The plane exists at a fixed distance $d$ from the origin, along the direction

$$ \hat{n} = \pmatrix{ \text{-}\cos \frac{\psi}{2} \\ \;\sin \frac{\psi}{2} \\ 0} $$

This defines the out of plane direction and also the two in-plane directions of $$ \hat{e} = \pmatrix{ \sin \frac{\psi}{2} \\ \cos \frac{\psi}{2} \\ 0} $$ and $\hat{k}$ which is common between the global coordinates and the plane coordinates.

We define a point in space $\boldsymbol{r}$ to be on the plane if it expressed as

$$ \boldsymbol{r} = d\,\hat{n} + u\,\hat{k} + v\,\hat{e} \tag{1}$$

where $(u,v)$ are the in-plane coordinates of the point. You can prove the above as it obeys the vector plane equation of $\boldsymbol{r} \cdot \hat{n} = d$ given a plane unit normal vector $\hat{n}$ and distance $d$ from the origin ($\cdot$ is the vector dot product).

Now these points must also obey the equation of a cone, which in this case can be written as

$$ \boldsymbol{r} \cdot \hat{j} = \| \boldsymbol{r} \| \cos \tfrac{\psi}{2} \tag{2}$$

The above states that the angle between the vector $\boldsymbol{r}$ and $\hat{n}$ should equal to $\psi/2$ for all $\boldsymbol{r}$.

The we can use the plane points from (1) in the cone equation (2) to get

$$ \begin{gathered}\left(d\,\hat{n}+u\,\hat{k}+v\,\hat{n}\right)\cdot\hat{j}=\|d\,\hat{n}+u\,\hat{k}+v\,\hat{n}\|\,\cos\tfrac{\psi}{2}\\ d\,\left(\hat{n}\cdot\hat{j}\right)+u\,\left(\hat{k}\cdot\hat{j}\right)+v\,\left(\hat{n}\cdot\hat{j}\right)=\sqrt{d^{2}+u^{2}+v^{2}}\,\cos\tfrac{\psi}{2}\\ v\cos\tfrac{\psi}{2}+d\sin\tfrac{\psi}{2}=\sqrt{d^{2}+u^{2}+v^{2}}\,\cos\tfrac{\psi}{2}\\ v+d\,\tan\tfrac{\psi}{2}=\sqrt{d^{2}+u^{2}+v^{2}}\\ \left(v+d\,\tan\tfrac{\psi}{2}\right)^{2}=d^{2}+u^{2}+v^{2}\\ \left(d\,\tan\tfrac{\psi}{2}\right)^{2}+2\left(d\,\tan\tfrac{\psi}{2}\right)\,v=u^{2}+d^{2}\\ v=\frac{u^{2}+d^{2}-\left(d\,\tan\tfrac{\psi}{2}\right)^{2}}{2\left(d\,\tan\tfrac{\psi}{2}\right)} \end{gathered}$$

which is clearly a parabola since it is of the form $v = a u^2 + b$ with coefficients $a = \tfrac{1}{2 d} \cot \tfrac{\psi}{2}$ and $b = d \cot \psi$ which are constants derived from the geometry of the problem.