Bézier curve approximating the intersection between a cone and a plane.

calculusconic sectionscurvesgeometry

I've been trying to fine a nice way to visualise projecting a parabola to a circular arc, but I realise I don't actually know how to derive the functions necessary to show projecting a parabola through its plane onto the base plane of the conic it lies on…

I started with the following geometric setup:

a parabolic conic section

In this, we have a 3D space with an origin $P_2=(0,0,0)$, and a cone with base radius $r$ and height $h$, effecting a conic angle $\phi$. There is a plane at the same angle, shifted along the x-axis by $d$, and as such the intersection of the plane and the cone is a parabolic conic section, with points $A$ and $C$ the intersection points of the plane and the conic base and point $B$ the parabola's extremum.

From the geometry, it is clear that every point on the parabola can be projected from $P_2$ through the parabolic plane onto the conic base plane to yield a circular arc (after all, the cone itself is that projection) but I have no idea how to turn this into the form of a function.

I figured I could represent the parabola as a 3D rational quadratic Bezier curve, with start point $A$ and end point $C$, and a control point $B'=\{ 0, r-d, \frac{r-d}{r}h \}$, and then vary the rational weight for point $B$ to "slide" the parabola along the cone, but that does not appear to be what happens when you adjust that value.

So my question is: given this geometry, and given the parabola as a Bezier curve, how would I go about turning that into a "projective" Bezier curve function that takes the coordinates $A$, $B$, and $C$, and the angle $\phi$, yielding the conic section described by the intersection of the cone and a plane passing through $A-C$ at angle $\phi$?

(E.g. at $\phi=0$ this would be a circular arc, at $\phi=\frac{\pi}{2}$ a triangle, and anywhere in between some form of non-parabolic elliptical arc)

Best Answer

One reason why your idea might not have panned out is that the point $B'$ in your question is the parabola’s vertex, so it doesn’t even work as the control point for a Bézier parameterization of the original parabola: even for a rational Bézier curve the middle control point never lies on the curve. The correct middle control point for a Bézier parameterization of the parabola can be found by intersecting the tangent planes to the cone at $A$ and $C$ with the plane that generates the parabola. We can thus find that with $$A = \left(2\sqrt{d(r-d)},r-2d,h\right) \\ B = \left(0,r,\frac hr(r-2d)\right) \\ C = \left(-2\sqrt{d(r-d)},r-2d,h\right)$$ the Bézier curve $$(1-t)^2A+2t(1-t)B+t^2C \tag1$$ matches the parabola. For $0\le t\le1$, we get the arc through the vertex with endpoints $A$ and $C$. (This orientation might not match your diagram exactly.)

Going back to your idea of projecting the parabola onto a variable plane through $A$ and $C$, we observe that since the center of the projection is the origin, if a point $P$ has a finite projection onto the plane then it is a scalar multiple of $P$ and so can be expressed as $P'=c(P)P$. Note that this scalar $c$ depends on $P$, and, of course, on the image plane. By construction, $A$ and $C$ are their own images, so the projection of the curve (1) is $$(1-t)^2A+2t(1-t)c(B)B+t^2C. \tag2$$ Comparing the rational Bézier parameterization $${(1-t)^2w_1P_1+2t(1-t)w_2P_2+t^2w_3P_3 \over (1-t)^2w_1+2t(1-t)w_2+t^2w_3}$$ to (2) suggests setting $w_1=w_3=1$ and since we’d like the control point to lie on the curve’s plane, $w_2=1/c(B)$ seems like a reasonable guess, leading to $${(1-t)^2A + 2t(1-t)B'/c(B)+t^2C \over (1-t)^2+2t(1-t)/c(B)+t^2}. \tag3$$

Parameterizing the variable plane by the angle $\phi$ between its normal and the positive $z$-axis, we have for an equation of the plane $$y\sin\phi+z\cos\phi=h\cos\phi+(r-2d)\sin\phi$$ and we can find $$B' = {r(h\cos\phi+(r-2d)\sin\phi) \over h(r-2d)\cos\phi+r^2\sin\phi}B.$$ I’ll leave verifying that using these values of $B'$ and $w$ do indeed produce a parameterization of the intersection curve, which involves some tedious but straightforward algebra.

Observe that, instead of sliding the vertex along the cone as you proposed, we are effectively sliding the parabola’s control point $B$ along the line through it and the origin. Parameterizing this family of curves by $c(B)$ as in (3) instead of the plane’s normal angle makes this relationship more transparent, I think.

Here, for example, is what results with $h=3$, $r=1$, $d=1/4$ and $\phi=\pi/4$:

The black arc is the above rational Bézier curve for $0\le t\le1$.

The locus

After doing some computations in the spirit (and building on the code) of my vertex anser, I found the locus of $P$ to be a quadric

$$ a\,x^2 + b\,y^2 + c\,xy + d\,x + e\,y + f = 0 $$

with the following parameters:

\begin{align*} a&=4\left(-A_y\,B_y + B_y^2 + A_y\,C_y - B_y\,C_y\right)\\ b&=4\left(-A_x\,B_x + B_x^2 + A_x\,C_x - B_x\,C_x\right)\\ c&=4\left(A_y\,B_x + A_x\,B_y - 2\,B_x\,B_y - A_y\,C_x + B_y\,C_x - A_x\,C_y + B_x\,C_y\right)\\ d&=4\left(A_y\,B_x\,B_y - A_x\,B_y^2 + A_y\,B_y\,C_x - B_y^2\,C_x - 2\,A_y\,B_x\,C_y + A_x\,B_y\,C_y + B_x\,B_y\,C_y\right)\\ e&=4\left(-A_y\,B_x^2 + A_x\,B_x\,B_y + A_y\,B_x\,C_x - 2\,A_x\,B_y\,C_x + B_x\,B_y\,C_x + A_x\,B_x\,C_y - B_x^2\,C_y\right)\\ f&=A_y^2\,B_x^2 - 2\,A_x\,A_y\,B_x\,B_y + A_x^2\,B_y^2 - 2\,A_y^2\,B_x\,C_x + 2\,A_x\,A_y\,B_y\,C_x \\&\quad{} - 2\,A_y\,B_x\,B_y\,C_x + 2\,A_x\,B_y^2\,C_x + A_y^2\,C_x^2 - 2\,A_y\,B_y\,C_x^2 + B_y^2\,C_x^2 \\&\quad{} + 2\,A_x\,A_y\,B_x\,C_y + 2\,A_y\,B_x^2\,C_y - 2\,A_x^2\,B_y\,C_y - 2\,A_x\,B_x\,B_y\,C_y \\&\quad{} - 2\,A_x\,A_y\,C_x\,C_y + 2\,A_y\,B_x\,C_x\,C_y + 2\,A_x\,B_y\,C_x\,C_y - 2\,B_x\,B_y\,C_x\,C_y \\&\quad{} + A_x^2\,C_y^2 - 2\,A_x\,B_x\,C_y^2 + B_x^2\,C_y^2 \end{align*}

Assumption confirmed

Using this formulation, I could verify your assumption: the conic is indeed a hyperbola, with the lines $AB$ and $CB$ as asymptotes. The way to check this is by ensuring that the lines are tangents, and that they touch the conic at infinity.

Special location

In a comment below, you mention that you might assume $B_x=B_y=0$. With that your conic will become

\begin{align*} a&=4\,A_y\,C_y\\ b&=4\,A_x\,C_x\\ c&=-4\left(A_y\,C_x + A_x\,C_y\right)\\ d&=0\\ e&=0\\ f&=A_y^2\,C_x^2 - 2\,A_x\,A_y\,C_x\,C_y + A_x^2\,C_y^2 \end{align*}

The formula is indeed a lot easier, so it might be a good starting point for a geometric description of the curve. For example, it is readily apparent that this hyperbola will be symmetric around the origin. But we already knew $B$ to be the center due to the asymptotes. You can even take this one step further and, via an affine transformation, consider only the case where

\begin{align*} A &= \begin{pmatrix}1\\0\end{pmatrix} & B &= \begin{pmatrix}0\\0\end{pmatrix} & C &= \begin{pmatrix}0\\1\end{pmatrix} \end{align*}

You end up with the hyperbola

$$4xy=1$$

From this you can e.g. conclude that the midpoint between $A$ and $C$ will also lie on that hyperbola. It will of course correspond to a Bézier curve which passes through $B$ for some $t<0$ or $t>1$. You can take that midpoint and reflect it in $B$ to obtain a point on the segment of the hyperbola you get for $0\le t\le 1$. This construction is invariant under affine transformations, so it still holds for the general case. Two asymptotes plus one point on the hyperbola amounts to five real degrees of freedom, so this should be enough to uniquely define your hyperbola.

Summary

To sum it up: the locus of the apex $P$ is the unique hyperbola with asymptotes $AB$ and $CB$ which passes through the midpoint between $A$ and $C$. The portion of it which corresponds to parabolas where $B$ lies between $A$ and $C$, i.e. is obtained in the Bézier curve for $0<t<1$, is the component of the hyperbola which does not contain that midpoint. It does contain the point obtained by reflecting that midpoint in $B$.

The code

Here is the sage code I used to obtain this representation:

# Define multivariate polynomial ring and points
PR1.<A_x, A_y, B_x, B_y, C_x, C_y, P_x, P_y, t> = QQ[]
A = vector(PR1, [A_x, A_y, 1])
B = vector(PR1, [B_x, B_y, 1])
C = vector(PR1, [C_x, C_y, 1])
P = vector(PR1, [P_x, P_y, 1])

# Quadratic Bézier curve parametrized by t
Bt = (1-t)^2*A + 2*(1-t)*t*P + t^2*C
r1 = (Bt[0] - B_x).resultant(Bt[1] - B_y, t) # eliminate t

# Obtain coefficients for coordinates of P
c1 = vector(PR1, flatten([list(i.polynomial(i.parent()(P_y)))
                          for i in r1.polynomial(P_x)]))
f, e, b, d, c, a = c1

# Print result
fmt1 = [str(i/4).replace('*','\\,')
        for i in [a, b, c, d, e]] + [str(f)]
fmt2 = [i + '&=4\\left(' + j + '\\right)'
        for i, j in zip('abcdef', fmt1)]
fmt2[-1] = 'f&='+fmt1[-1]
print('\\\\\n'.join(fmt2))

# Check whether lines AB and CB are asymptotes of the Hyperbola
Hyperbola = Matrix([
  [2*a, c, d],
  [c, 2*b, e],
  [d, e, 2*f]])
def onConic(p, c=Hyperbola):
    return (p.row()*c*p.column())[0,0].is_zero()
asymptotes = [B.cross_product(p) for p in [A, C]]
infLine = vector(QQ, [0,0,1])
# asymptotes are tangents to the hyperbola:
assert(all(onConic(i, Hyperbola.adjoint()) for i in asymptotes))
# asymptotes touch the hyperbola at infinity:
assert(all(onConic(i.cross_product(infLine)) for i in asymptotes))

# The midpoint between A and C is on the hyperbola:
assert(onConic(A+C))

Originally I had more complicated code which did not rely on the interpretation as a Bézier curve. The result was the same, though.

Generalizing to rational case

Can we generalize to rational quadratic Bezier curves (i.e. to conic section curves that are not necessarily parabolas)?

A parabola has four real degrees of freedom. If you choose a non-rational Bézier curve, you have two real degrees of freedom, but with these you not only specify a parabola but also select a start point and an end point on that parabola, so the degrees of freedom match. A conic in general has five real degrees of freedom. So if you want it to pass through three given points, that still leaves a two-parameter family of corresponding conics. Therefore your locus will not be a single curve, but either the whole plane or some portion of it.

You can define a conic using five points through which it should pass. In addition to your $A,B,C$ you might use two more points, which you move close to the end points $A$ and $C$. By making them arbitrary close (i.e. computing some limit), you can use these control points to exactly and arbitrarily determine the direction of the tangents in $A$ and $C$. Therefore you can choose any point $P$ in the plance, and find a conic through $A,B,C$ which will have $P$ as its apex. In this sense, the whole plane will be your locus.

I'm not completely sure whether a rational Bézier curve can be defined in such a way that it passes through infinity, but I believe that to be the case. If not, then there might be cases where the resulting conic would be a hyperbola, and $A,B,C$ are not all three in the same of its components. This might result in a restriction to part of the plane.

[Math] Questions on the relation of the axis of a cone to its conic sections

Converting comments to answer, as requested:

The Dandelin spheres answer question (1): a focus of a conic section is the point of tangency of its plane with one of those spheres. Clearly, the point on tangency lies on the cone axis if and only if the plane is perpendicular to that axis; therefore, the axis contains a focus in, and only in, the case of a circle. The axis point has some relation to the conic section, but not one as interesting or useful as a focus. (As for "Does the diameter of the base of the cone pass through the focus of the hyperbola?": note that the cone has no base. The thing extends, and expands, infinitely-far.)

This notion of points seeming to move "out to infinity and come back on the other side" is not uncommon in analytical geometry. (It's kinda what happens along asymptotes, when you think about it.) You might be interested in studying Projective Geometry, which adds a "line at infinity" to the standard Euclidean plane; this broader context helps unify all the conic sections into a single kind of curve that has various different relations to that line. (The parabola, in particular, has its "second vertex" on that line.)