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I saw this post recently deriving the equation of a conic section from the equation of a plane and a double cone.

I'm attempting the inverse, solving for the equation of a plane given the double cone and conic section.

Following the logic of the linked post, we have the following definitions:

$$p'x + q'y + z + s' = 0$$ $$ax^2 + by^2 + cz^2 + dxy + exz + fyz = 0$$ $$A = a + cp'^2 - ep'$$ $$B = d + 2cp'q' - eq' - fq'$$ $$C = b + cq'^2 - fq'$$ $$D = 2cp's' - es' = s' (2cp' - e)$$ $$E = 2cq's' - fs' = s' (2cq' - f)$$ $$F = cs'^2$$ $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$

Solving for $s'$ is easy since we have the F equation (which we can then substitute into other equations for $s'$): $$s'=\sqrt{\frac{F}{c}}$$ Solving the equations of $A$ and $D$ for $0$, we get the following equation: $$cp'^2 - ep' + a - A - 2cp's' - es' = 0$$ Which can then be solved for $p'$ (and checked with Wolfram|Alpha): $$p'=\frac{e+2c\sqrt{\frac{F}{c}}\pm\sqrt{e^2-4c(D+a-A-F)}}{2c}$$

The same logic when solving for $p'$ can be used to solve for $q'$: $$q'=\frac{f+2c\sqrt{\frac{F}{c}}\pm\sqrt{f^2-4c(E+b-C-F)}}{2c}$$

I plugged it all into GeoGebra's 3D calculator: https://www.geogebra.org/3d/t7feqgza

At first, everything seemed fine, though upon changing coefficients I found the graph is undefined.

For example, if the sign of $a$ and $b$ are the same as the sign of $c$, the plane is undefined. Furthermore, if $d$, $e$, and $f$ are all equal to $0$ (in addition to the other double cone coefficients having the same sign) then the conic is undefined.

My only possible explanation for this is that the $B$ equation is not used when deriving the equations of $p'$ and $q'$. The next step would obviously be to derive the equations of $p'$ and $q'$ including the $B$ equation. Except for that the $B$ equation includes both $p'$ and $q'$ so neither of them can be solved without including the other, which has the same problem in turn.

What are the equations for $p'$ and $q'$?

Edit #1

After receiving some negative feedback, I'd like to clarify a few things about the problem:

  1. The values of $a, b, c, d, e, f, A, B, C, D, E$ and $F$ are known. In addition, the following condition is true: $$\{a,b,c,d,e,f,A,B,C,D,E,F\} \subset\mathbb{R}$$
  2. The double cone is defined by $$ax^2+by^2+cz^2+dxy+exz+fyz=0$$
  3. The conic section is defined by $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
  4. Both the double cone and plane are infinite (so there is only one possible plane)

Edit #2 For more clarification:

  1. In the equation of the conic section, a point $P(x,y)$ lies on the plane.

  2. In the equation of the plane and the double cone, a point $P(x,y,z)$ is in 3D coordinate space.

Jim
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    I solve this question in a different way, it involves finding the right circular cones that intersect $z=0$ in the given conic section and translating and rotating $z=0$ to get the plane you are seeking. Do you want to see this solution? – Jan-Magnus Økland Feb 14 '23 at 06:53
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    E.g. $x^2-xy+y^2-1=0$ has a right circular cone intersecting it with $z=0$ with vertex $(-2\sqrt2/3,-2\sqrt2/3,-\sqrt2/3)$ and unit direction of axis $[\frac1{\sqrt6},\frac1{\sqrt6},\frac2{\sqrt6}].$ Now the plane you seek is the image of $z=0$ by translating $(-2\sqrt2/3,-2\sqrt2/3,-\sqrt2/3)$ to the origin and rotating $[\frac1{\sqrt6},\frac1{\sqrt6},\frac2{\sqrt6}]$ to $[0,0,1].$ – Jan-Magnus Økland Feb 14 '23 at 08:22
  • @Jan-MagnusØkland could you post an answer with the process you use? Also note it should be a linear algebra answer. – Jim Feb 14 '23 at 15:19
  • @Intelligentipauca For this case both shapes (the double cone and the plane) are infinite -- there should only be one possible plane that intersects the given double cone to result in the given conic section. – Jim Feb 15 '23 at 19:47
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    You are right, sorry: the aperture of the cone is fixed. Here's a geometric way to solve the problem: https://math.stackexchange.com/a/2351900/255730 – Intelligenti pauca Feb 15 '23 at 20:59
  • @Intelligentipauca I'm not entirely sure why it's such a hard problem to solve. We have 3 unknowns and 9 equations. As a system of equations, it should be easy to solve. – Jim Feb 16 '23 at 14:06
  • In addition, various aspects of the question make it easier to solve. – Jim Feb 16 '23 at 14:08
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    You should make clear what coordinates are used for the conic section: is that a system of coordinates in the plane? And how is it chosen? – Intelligenti pauca Feb 16 '23 at 14:19
  • @Intelligentipauca I've added a couple of edits for clarification. – Jim Feb 16 '23 at 14:23
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    You didn't make clear what coordinates are used in the conic section equation. As written, it is the equation of a cylinder. If so, you'd better point it out. Otherwise please explain what is the meaning of that equation. – Intelligenti pauca Feb 16 '23 at 14:42
  • @Intelligentipauca The equation of the conic section are a 2D shape (no z at all), not 3D. Cylinders do not exist in 2D space. – Jim Feb 16 '23 at 14:52
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    Of course. But the cone is in 3D space and so is the plane where the conic section lies. HOW ARE COORDINATES SET UP IN THAT PLANE? – Intelligenti pauca Feb 16 '23 at 15:08
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    Unless your equation doesn't represent the conic section itself, but its projection on the x-y plane. Is it so? – Intelligenti pauca Feb 16 '23 at 15:14
  • @Intelligentipauca yes, the Conic section equation is that of the shape projected onto the plane. – Jim Feb 18 '23 at 21:20
  • Fine. Then the solution is simple, in principle at least: the plane is where the intersection of the cone with the (non-circular) cylinder lies. The problem is that, in general, the intersection of a cone with a cylinder DOES NOT LIE on a plane. – Intelligenti pauca Feb 18 '23 at 22:00
  • See here, for instance, the intersection of cone $x^2+y^2={1\over3}z^2$ with cylinder $^2−+^2=1$: https://i.sstatic.net/5q30K.png – Intelligenti pauca Feb 18 '23 at 22:15
  • The conic section is a 2D shape, not 3D. Take for example this conic section: https://www.desmos.com/calculator/4nggnvydyf . The shape shown is the edge of intersection of a double cone and the plane (drawn on the plane). – Jim Feb 19 '23 at 22:23
  • $x+y=0$ where you also have $z$ as you do, is a plane not a line. It's the same with degree $2$ equations. To specify a curve in 3d, you need more than one equation: $x+y=0,z=0,$ say. For the conic section, you need both the cone and the plane. You provide an equation (only implicitly) in the plane $z=0$ that would cut @Intelligentipauca s cone over it in the 2d conic. Or a right circular cone, like the one I provide above, that cuts out the same conic in $z=0$. But you fix the cone and want a plane, that cuts out the conic, rigidly moved from the $z=0$ plane, onto this other plane, right? – Jan-Magnus Økland Feb 20 '23 at 12:29

1 Answers1

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The method is here only change apex angle to $\frac{\pi}{4}.$ I've put it in a maxima CAS script

rcc(eq1,i):=block([a,b,c,d,e,f,cx,cy,cz,n1,n2,n3,A,B,C,D,k],
a:coeff(eq1,x^2),
c:coeff(eq1,y^2),
f:coeff(coeff(eq1,x,0),y,0),
b:coeff(coeff(eq1,x),y),
d:coeff(coeff(eq1,x),y,0),
e:coeff(coeff(eq1,y),x,0),
m:matrix([a,b/2,A/2,d/2],[b/2,c,B/2,e/2],[A/2,B/2,C,D/2],[d/2,e/2,D/2,f]),
s:solve([determinant(m)],[C]),
C:rhs(s[1]),
eq2:eq1+z*(A*x+B*y+C*z+D),
s2:solve([diff(eq2,x),diff(eq2,y),diff(eq2,z)],[x,y,z]),
cx:rhs(s2[1][1]),
cy:rhs(s2[1][2]),
cz:rhs(s2[1][3]),
eq3:ratexpand(a*(x+cx)^2+b*(x+cx)*(y+cy)+c*(y+cy)^2+d*(x+cx)+e*(y+cy)+f+(z+cz)*(A*(x+cx)+B*(y+cy)+C*(z+cz)+D)),
eq4:ratexpand(eq3-k*((1/2)*(x^2+y^2+z^2)-(n1*x+n2*y+n3*z)^2)),
s3:solve([coeff(eq4,x,2),coeff(eq4,y,2),coeff(eq4,z,2),coeff(coeff(eq4,x),y),coeff(coeff(eq4,x),z),coeff(coeff(eq4,y),z),n1^2+n2^2+n3^2-1],[A,B,D,n1,n2,n3,k]),
A:rhs(s3[i][1]),
B:rhs(s3[i][2]),
C:'C,
D:rhs(s3[i][3]),
n1:rhs(s3[i][4]),
n2:rhs(s3[i][5]),
n3:rhs(s3[i][6]),
m:matrix([a,b/2,A/2,d/2],[b/2,c,B/2,e/2],[A/2,B/2,C,D/2],[d/2,e/2,D/2,f]),
s4:solve([determinant(m)],[C]),
C:rhs(s4[1]),
eq6:eq1+z*(A*x+B*y+C*z+D),
s4:solve([diff(eq6,x),diff(eq6,y),diff(eq6,z)],[x,y,z]),
cx:rhs(s4[1][1]),
cy:rhs(s4[1][2]),
cz:rhs(s4[1][3]),
return([ratsimp(eq6),n1,n2,n3,cx,cy,cz]));

rcc returns a list of a right circular cone intersecting $z=0$ in the given conic, components of unit direction of the axis and cone point for the $i$-th solution. Some may not be real and sometimes maxima can't find the solution without a grobner basis computation.

To solve your problem, we can translate and rotate $$(1/2)((x-c_x)^2+(y-c_y)^2+(z-c_z)^2)-(n_1(x-c_x)+n_2(y-c_y)+n_3(z-c_z))^2)=0$$ to $x^2+y^2-z^2=0$ and this rigid motion will take $z=0$ to the plane you're seeking. I.e. the plane $$n_1(x-c_x)+n_2(y-c_y)+n_3(z-c_z)=0$$ is mapped to $z=0$ by this rigid motion.

The cone and two planes

Jan-Magnus Økland
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