Properties of a hyperbola slice of a cone

I'm attempting to locate some points (in a 3D coordinate system) on the surface of a cone by slicing the cone with a plane, and using the resultant ellipse, parabola or hyperbola to calculate the points. I've had success with the ellipse & parabola solutions, but am seeking help with the hyperbola case.

Known data: The slope of the cone, the slope of the plane, and distance from the vertex of the cone to the intersection of the plane and cone's longitudinal axis. Given these knowns, I'm looking for "a" & "b".

From reading, I have the formulas x²/a² - y²/b² = 1 & c² = a² + b². I understand distance "a" to be half the distance between the two vertices. Question 1 is, is "a" always half the distance between the two vertices, even when the plane is sloped? If not always half, how is "a" calculated when the plane is sloped? Question 2 is how to determine "b"? Question 3 is how to determine the angle of the asymptotes? ...is that the angle of the cone, or if the plane is sloped, some other angle?

I'm concerned only with the upper nappe of the cone.

As I massage the above formula, attempting to solve for an "x" & "y" & "b", as the plane angle gets shallower, "a" gets longer. And when "a" gets longer than an "x" value that I am using, my solutions fail due to an attempt to take the square root of a negative number. So, I know I'm messing up somewhere. Any help is appreciated.

This is my first post here so I don't know my way around. I'm going to attempt to post a sketch.

• This is interesting. I will work on this problem when I get home. Can we assume this is a right-angled cone? Or is the interior angle something else? Commented Jan 2, 2021 at 19:11
• Thank you. It is a right cone. Commented Jan 2, 2021 at 19:13
• see Dandelin Sphere construction. "A conic section has one Dandelin sphere for each focus. An ellipse has two Dandelin spheres touching the same nappe of the cone, while hyperbola has two Dandelin spheres touching opposite nappes. A parabola has just one Dandelin sphere. " en.wikipedia.org/wiki/Dandelin_spheres Commented Jan 2, 2021 at 19:21
• @WoodGuy This problem is turning out to be harder than I expected. Hopefully I will post something by the end of today. Commented Jan 3, 2021 at 20:31
• I have now posted a partial answer. Commented Jan 4, 2021 at 3:22

Here's the plan of attack.

• Give equations that describe the cone and plane in the standard basis

• Define "plane coordinates"

• Find the coordinate transform matrix between the two coordinate systems

• Find the equation of the conic section in plane coordinates

• Rotate the coordinate system so our conic is oriented along the coordinate axes and thus find the major and minor axis lengths

Once we have the equation for the conic, there will still be a lot of work before we can get the semi major and semi minor axes, but hopefully someone else can chime in on this matter. First a preface on notation - scalars will be denoted with normal italic font, e.g $$s$$. Vectors will be denoted with upright (Roman) font, e.g $$\mathrm{v}$$. Matrices will be denoted with bold, e.g $$\mathbf{M}$$. With that out of the way, let's get started!

1: Cartesian equations of the plane and cone

First, the plane. Any plane in $$\Bbb{R}^3$$ is a level surface of a linear polynomial in $$x,y,z$$. What I mean by this is given $$a,b,c\in\Bbb{R}$$ we can represent the corresponding plane as $$\mathcal{P}=\{(x,y,z)\in\Bbb{R}^3:ax+by+cz=d\}$$ Yes, apologies on using $$a$$ and $$b$$ - I will probably use them again later.

Now, the cone. The upward facing right angled cone is the surface corresponding to $$\phi=\pi/4$$, in spherical coordinates. The inverse transformation tells us that this means $$\arccos\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)=\frac{\pi}{4}$$ Taking the cosine of both sides we see that $$\frac{z^2}{x^2+y^2+z^2}=\frac{1}{2}$$ So we get $$\mathcal{C}=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2-z^2=0\}$$ Now while we initially only wanted the upward part of the cone, we can see that rewriting this as $$z^2=\pm\sqrt{x^2+y^2}$$ Means that our equation gives the lower part of the cone as well. So as long as our plane isn't vertical, i.e $$c=0$$, the system of equations $$\begin{bmatrix} ax+by+cz\\ x^{2} +y^{2} -z^{2} \end{bmatrix} =\begin{bmatrix} d\\ 0 \end{bmatrix}$$ Which will always have at least one solution.

2: Plane coordinates

Our goal here is to create a new "basis" (this is technically an abuse of terminology) for $$\mathbb{R}^3$$ that we can write a vector as an affine transformation $$x\hat{\mathrm{i}}+y\hat{\mathrm{j}}+z\hat{\mathrm{k}}=(x'\hat{\mathrm{i}}'+y'\hat{\mathrm{j}}'+z'\hat{\mathrm{k}}')+\mathrm{p}$$ Where $$\mathrm{p}$$ is some reference point on the plane, and the primed and unprimed basis vectors are related through via a linear transformation $$\mathbf{M}\begin{bmatrix} \hat{\mathrm{i}}\\ \hat{\mathrm{j}}\\ \hat{\mathrm{k}} \end{bmatrix} =\begin{bmatrix} \hat{\mathrm{i}}'\\ \hat{\mathrm{j}}'\\ \hat{\mathrm{k}}' \end{bmatrix}$$ The inverse transformation between coordinate systems is $$\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\mathbf{M}^{-1}\begin{bmatrix} x'\\ y'\\ z' \end{bmatrix}+\mathrm{p}$$ We can multiply both sides by $$\mathbf{M}$$ to get the forward transformation: $$\mathbf{M}\left(\begin{bmatrix} x\\ y\\ z \end{bmatrix}-\mathrm{p}\right)=\begin{bmatrix} x'\\ y'\\ z' \end{bmatrix}$$ First, we find the point $$\mathrm{p}$$. Though any point will do, it makes sense to use the one closest to the origin (of our original coordinate system). It turns out that this point is $$\mathrm{p}=\frac{d}{a^2+b^2+c^2}(a,b,c)$$ This can be obtained either with simple coordinate geometry or, as I did it, with Lagrange multipliers. I can elaborate on how I did this if you wish. We let our third new basis vector be a unit vector parallel to this vector, that is $$\hat{\mathrm{k}}'=\frac{(a,b,c)}{\sqrt{a^2+b^2+c^2}}$$ We want our new basis to be orthonormal, that is we want the other two basis vectors to satisfy the equations $$\hat{\mathrm{i}}'\boldsymbol{\cdotp}\hat{\mathrm{k}}'=\hat{\mathrm{j}}'\boldsymbol{\cdotp}\hat{\mathrm{k}}'=\hat{\mathrm{i}}'\boldsymbol{\cdotp}\hat{\mathrm{j}}'=0$$

3: Finding the precise transformation matrix between the unit vectors

As of now, we have three equations for six (three components for both i and j) unknowns, so we need three extra conditions so we can uniquely determine our coordinate system. We'll only impose two extra conditions, namely $$i'_1=j'_2$$ and $$j'_1=i'_2$$. Letting $$i'_1=1$$ for now (this will be normalized later) we get the solutions $$\hat{\mathrm{i}}'\propto \left(1,\frac{-a^2-b^2-2c^2+\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2ab},\frac{-a^2+b^2+2c^2-\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2ac}\right)$$ $$\hat{\mathrm{j}}'\propto \left(\frac{-a^2-b^2-2c^2+\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2ab},1,\frac{a^2-b^2+2c^2+\sqrt{(a^2+b^2+2c^2)^2-4a^2b^2}}{2bc}\right)$$ For first one I'll use the abbreviations $$\mu,\nu_1$$ (in that order) and for the second one I'll use the abbreviations $$\mu,\nu_2$$ in that order. I'll also let $$\kappa_1=\sqrt{1+\mu^2+\nu_1^2}$$ and $$\kappa_2=\sqrt{1+\mu^2+\nu_2^2}$$ so we can normalize the above two expressions and write $$\hat{\mathrm{i}}'=\frac{1}{\kappa_1}(1,\mu,\nu_1)~~;~~\hat{\mathrm{j}}'=\frac{1}{\kappa_2}(\mu,1,\nu_2)$$ The gammas can be found exactly in terms of $$a,b,c$$, but the expressions are rather lengthy so I won't bother posting them here. In practice you would do this numerically, not symbolically, anyway. Regardless, from the above expressions it's clear that $$\hat{\mathrm{i}}'=\frac{1}{\kappa_1}(\hat{\mathrm{i}}+\mu\hat{\mathrm{j}}+\nu_1\hat{\mathrm{k}})~~;~~\hat{\mathrm{j}}'=\frac{1}{\kappa_2}(\mu\hat{\mathrm{i}}+\hat{\mathrm{j}}+\nu_2\hat{\mathrm{k}})$$ And of course $$\hat{\mathrm{k}}'=\frac{a}{\sqrt{a^2+b^2+c^2}}\hat{\mathrm{i}}+\frac{b}{\sqrt{a^2+b^2+c^2}}\hat{\mathrm{j}}+\frac{c}{\sqrt{a^2+b^2+c^2}}\hat{\mathrm{k}}$$ Hence we can write the inverse transformation as $$\begin{bmatrix} \hat{\mathrm{i}}'\\ \hat{\mathrm{j}}'\\ \hat{\mathrm{k}}' \end{bmatrix} =\underbrace{\begin{bmatrix} \frac{1}{\kappa _{1}} & \frac{\mu }{\kappa _{1}} & \frac{\nu _{1}}{\kappa _{1}}\\ \frac{\mu }{\kappa _{2}} & \frac{1}{\kappa _{2}} & \frac{\nu _{2}}{\kappa _{2}}\\ \frac{a}{\sqrt{a^{2} +b^{2} +c^{2}}} & \frac{b}{\sqrt{a^{2} +b^{2} +c^{2}}} & \frac{c}{\sqrt{a^{2} +b^{2} +c^{2}}} \end{bmatrix}}_{\mathbf{M}}\begin{bmatrix} \hat{\mathrm{i}}\\ \hat{\mathrm{j}}\\ \hat{\mathrm{k}} \end{bmatrix}$$ We can, of course, invert this matrix to get the forward transformation, but the expressions become extremely long, so I'll once again omit them. Recalling the inverse transformation, points $$(x,y,z)$$ sitting on the plane will have a zero $$z'$$ coordinate, precisely, they will be expressible as $$\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\mathbf{M}^{-1}\begin{bmatrix} x'\\ y'\\ 0 \end{bmatrix}+\mathrm{p}$$

4: Equation of the conic section

Recalling that the equation of the cone is $$x^2+y^2-z^2=0$$, we can plug in the RHS from the previous section into the equation for the cone to get the intersection surface between our plane and cone. We can represent our conic section as the quadric plane curve in our plane $$\mathcal{P}$$ using our primed plane coordinates $$\begin{array}{l} f(x',y')=\\ +\left( (\mathbf{M}^{-1} )_{1,1} \ x'+(\mathbf{M}^{-1} )_{1,2} \ y'+\frac{da}{a^{2} +b^{2} +c^{2}}\right)^{2}\\ +\left( (\mathbf{M}^{-1} )_{2,1} \ x'+(\mathbf{M}^{-1} )_{2,2} \ y'+\frac{db}{a^{2} +b^{2} +c^{2}}\right)^{2}\\ -\left( (\mathbf{M}^{-1} )_{3,1} \ x'+(\mathbf{M}^{-1} )_{3,2} \ y'+\frac{dc}{a^{2} +b^{2} +c^{2}}\right)^{2}\\ =0 \end{array}$$ So it seems computing the third row of the inverse matrix is actually not needed. However, numeric matrix inversion is very fast in software such as Mathematica, so I won't worry about it too much.

5: Reorienting our coordinate system

Let's drop the primes to make life easier for ourselves. We need to convert the above expression for $$f$$ into something of the form $$f(x,y)=A{x}^2+Bxy+C{y}^2+D=0$$ Which will be more useful for further analysis. We can avoid a pile of algebra (useful for numerical calculations) by simply noting $$\begin{array}{l} D=f( 0,0)\\ A=f( 1,0) -D\\ C=f( 0,1) -D\\ B=f( 1,1) -( A+C+D) \end{array}$$ Recall the standard (no tilting) equation for a conic is $$ux^2+vy^2-1=0$$ So the goal here is to find some change of coordinates $$(x,y)\to(X,Y)$$ that will eliminate the mixed $$XY$$ terms. We want to find some polynomial function $$F$$ such that $$f(x,y)=F(X,Y)\propto u X^2+vY^2-1$$ It turns out that a linear, not just affine, transformation will do - we don't need to move the origin. And we don't need to stretch the coordinate grid either, since our plane coordinates are already orthonormal. So we can simply use a rotation. So we are looking for a transformation of the form $$\begin{bmatrix} X\\ Y \end{bmatrix} =\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}$$ The inverse transformation is of course $$\begin{bmatrix} x\\ y \end{bmatrix} =\begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} X\\ Y \end{bmatrix}$$ Which means that $$\begin{array}{l} F(X,Y)=\\ A( X\cos \theta +Y\sin \theta )^{2}\\ +B( X\cos \theta +Y\sin \theta )( -X\sin \theta +Y\cos \theta )\\ +C( -X\sin \theta +Y\cos \theta )^{2}\\ +D \end{array}$$ So we expand out and set the mixed term equal to zero: $$\text{mixed coefficient}=2A\cos\theta\sin\theta+B(\cos^2\theta-\sin^2\theta)-2C\sin\theta\cos\theta=0$$ Using some double angle identities we get $$(A-C)\sin(2\theta)+B\cos(2\theta)=0$$ Obviously if $$B=0$$ this whole process is unnecessary and our original coordinates are already correctly aligned. Apart from that however, we break the solution of this equation into two cases.

Case 1 - $$A=C$$

When $$A=C$$, our equation reduces to
$$B\cos(2\theta)=0$$ So our first solution is $$\theta=\pi/4$$. Now since $$\sin(\pi/4)=\cos(\pi/4)$$ we can factor them out from our expression for $$F$$: $$F(X,Y)=\frac{1}{2}\left(C(X+Y)^2+B(Y+X)(Y-X)+C(X-Y)^2+2D\right)$$ Doing some algebra, we get $$F(X,Y)=\left(C-\frac{B}{2}\right)X^2+\left(C+\frac{B}{2}\right)Y^2+D$$ So $$F(X,Y)=-D\left(\underbrace{\frac{B-2C}{2D}}_{\equiv u}X^2+\underbrace{\frac{-B-2C}{2D}}_{\equiv v}Y^2-1\right)$$ If either $$u=0$$ or $$v=0$$ then we have a parabola. If not, then if the signs of $$u$$ and $$v$$ are different, we have a hyperbola. If both are nonzero and the same sign, then we have an ellipse or circle, and $$"b"=\text{semi minor axis}=\beta=\sqrt{\frac{1}{\min(|u|,|v|)}}$$ $$"a"=\text{semi major axis}=\alpha=\sqrt{\frac{1}{\max(|u|,v|)}}$$ Meaning if $$|u|>|v|$$ our conic can be written in the form $$\pm \frac{x^2}{\alpha^2}\pm \frac{y^2}{\beta^2}=1$$ Which perhaps is more familiar. The alpha and beta go in opposite places if $$|u|<|v|$$.

Case 2: $$A\neq C$$

In this case the equation for $$\theta$$ is not as simple. Recall that we want $$(A-C)\sin(2\theta)+B\cos(2\theta)=0$$ We can restate this as $$\tan(2\theta)=\frac{B}{C-A}$$ So $$\theta=\frac{1}{2}\arctan\left(\frac{B}{C-A}\right)$$ Our mixed term is zero, by construction. The other coefficients are $$X\text{ squared coefficient}=A\cos^2\theta-B\cos\theta\sin\theta+C\sin^2\theta$$ $$Y\text{ squared coefficient}=C\cos^2\theta+B\cos\theta\sin\theta+A\sin^2\theta$$ So then $$u=\frac{A\cos^2\theta-B\cos\theta\sin\theta+C\sin^2\theta}{-D}$$ And $$v=\frac{C\cos^2\theta+B\cos\theta\sin\theta+A\sin^2\theta}{-D}$$ Once again if either is zero, then the figure we have is a parabola. If the signs are different, we have a hyperbola. If both are nonzero with the same sign, we have an ellipse or hyperbola, and the same equations hold: $$"b"=\text{semi minor axis}=\beta=\sqrt{\frac{1}{\min(|u|,|v|)}}$$ $$"a"=\text{semi major axis}=\alpha=\sqrt{\frac{1}{\max(|u|,|v|)}}$$

In response to comments - some examples with plots

Shown below is a graphic of the intersection between the plane $$ax+by+cz=d$$ with the cone $$x^2+y^2-z^2=0$$, using $$a=0.7,b=4.1,c=1,d=-3.3$$:

And now the plot of the conic of intersection $$Ax^2+Bxy+Cy^2+D=0$$, which we achieved through "plane coordinates" - in this case the coefficients are $$A=0.934531,B=-0.291272,C=-1.57866,D=0.530046$$:

Finally, we rotate our coordinate system by an amount $$\theta=0.0576914$$ to get the conic in the form $$ux^2+vy^2-1=0$$, in this case $$u=-1.71572,v=2.99421$$. The signs are different, hence we have a hyperbola - This is just a rotated version of the previous plot.

A quick recap

We started with the equations for the plane and the cone, $$ax+by+cz=d$$ and $$x^2+y^2-z^2=0$$. Now we have an equation for their intersection, which I have as $$ux^2+vy^2-1=0$$ Which perhaps you're more used to seeing as $$\pm\frac{x^2}{"a"^2}\pm\frac{y^2}{"b"^2}=1$$ The quotation marks are because I used a and b earlier and I don't want to cause any confusion.

Slope and angle

Let's assume now that $$u$$ and $$v$$ have different signs, i.e, we have a hyperbola. How can we find the equations of the asymptotes and the angle between them? First, the equations. We can solve the equation for $$y$$ to get $$y=\pm\sqrt{\frac{1-ux^2}{v}}$$ Which, since $$u$$ and $$v$$ are opposite signs, is $$y=\pm\sqrt{\frac{1}{v}+\frac{|u|}{|v|}x^2}$$ So for large $$x$$, the asymptotes are the lines $$y=\pm\sqrt{\frac{|u|}{|v|}}x$$ The line with the positive slope makes an angle of $$\arctan(\sqrt{|u|/|v|})$$ with the positive $$x$$ axis whereas the other makes an angle of $$\pi-\arctan(\sqrt{|u|/|v|})$$ with the positive $$x$$ axis meaning that the angle between them is $$\phi_1=\pi-2\arctan(\sqrt{|u|/|v|})$$. The other angle between them is $$\phi_2=\frac{2\pi-2\phi_1}{2}$$.

• This answer is incomplete as it stands, but I might get back to it sometime later. Hopefully this is a good start for now. Commented Jan 4, 2021 at 2:45
• K. I appreciate all the energy you put into this explanation. Please refer to Wikipedia on Conic Sections. Q1 - is "a" always half way? Or if the plane is sloped, what is solution to "a"? Q2: How to determine "b"? Q3: How do I determine the angle of the asymptotes if the plane is sloped? Commented Jan 4, 2021 at 6:56
• Sorry. Comments are limited to 230 characters so I had to abbreviate my response. Commented Jan 4, 2021 at 6:58
• @Woodguy Yes, indeed more work needs to be done to get to "a" and "b" and the angle between the hyperbola asymptotes. I have to go to work today, but I will update the answer with that info when I am finished. Commented Jan 4, 2021 at 13:36
• K. Your impressive and elaborate solution seems to be created by a quite talented mathematical mind, and presented in an extraordinarily professional manner, almost as if it should be published in a graduate level text of some variety. The only problem I can possibly find with it is... I don't get it. You lost me somewhere around "affine". I'm sorry. I'm simpler than that. And so is the solution to the 3 questions that I've posted twice, now one... three times. How to find "a"? And I suspect it doesn't take all that. Commented Jan 7, 2021 at 3:33

I did not see a diagram for the Dandelin Spheres with a hyperbola, so I drew one. This page has a hyperbola

This is a side view, a cross section. Each circle means a sphere, both spheres are tangent to the cone, which is shown by the two black lines. The red/purple line is (two generating lines of) the cone.

• Thanks Will. This provides the info to locate the focal point and gives me something to study further. Appreciate it. Commented Jan 2, 2021 at 20:56