I am able to follow the procedure of rewriting a general conic section in standard form but have a question concerning what is actually happening geometrically at each step. I will break these down into several parts:

1, The general equation of the conic section is $ax^2 + bxy + cy^2 + dx + ey + f =0$. How can we understand this as being a slice of a cone by a plane? When I google the picture of the various conic sections available, we have this double cone whose equation is presumably $X^2+Y^2=|Z|$ where I've introduced $(X,Y,Z)$ as coordinates on an ambient 3d space whilst $(x,y)$ are coordinates on the 2d plane. The equation of the plane in the ambient 3d space is going to be like $AX+BY+CZ=Q$. Now, if I substitute $Z=\frac{1}{C(Q-AX-BY)}$ into the cone equation and relabel the constants, I can get something like $X^2+Y^2+DX+EY+F=0$ but this has no $XY$ cross term, nor do I understand how to get it in terms of the $(x,y)$ coordinates that exist on the 2d planar surface (and thus that actually describe the conic section).

TLDR: Where does the general equation of the conic section come from? Can somebody convince me that it is indeed the intersection of a cone and a plane?

2, We can then rewrite $ax^2+bxy+cy^2+dx+ey+f=0$ as $\vec{x}^T A \vec{x} + \vec{v}^T \cdot \vec{x} + f=0$. My question here is what does each part describe? Intuitively I thought that $\vec{v}^T \cdot \vec{x}$ represents a translation of the cone within the ambient 3d space such that we get the same conic section but "off=centre" in terms of the 2d coordinates $(x,y)$. Is this true? If I have $b=0$ and keep the plane in the same place but move the cone then surely their intersection will not only have a different centre, but it will also have a different size i.e. it will be the same shape of section but either scaled up or down - do people agree with this? Then surely this term needs to describe more than just a translation but also a scaling - does this happen?

3, What about the case where $b \neq 0$. In this case, I have read that this represents a rotation of the conic section. What causes this - does the cone rotate or the plane? This is a part that I really struggle to get my head around since the matrix $A$ does not have unit determinant and therefore doesn't represent a rotation matrix, does it? Typically we would construct the matrix of orthonormal eigenvectors $P$ which would have unit determinant and use this to describe a rotation to a new coordinate system $(x',y')$ in which the conic section assumes the standard form. As I understand it, the symmetry of $A$ means that there will always exist such a $P$ that orthogonally diagonalises $A$ (spectral theorem) and thus $P$ describes a rotation from $(x,y)$ to $(x',y')$ where $x'$ and $y'$ point in the eigendirections of $A$ i.e. the eigendirections are the new major and minor axes for the conic section when written in $(x',y')$ coordinates. BUT if that's true, then P describes a rotation and A describes WHAT EXACTLY????

4, How can I understand why the eigendirections of $A$ will be the major and minor axes of the conic section in the $(x',y')$ coordinates?

THank you very much for your help. I am just struggling to visualise what's actually happening behind the relatively straightforward mathematics here.


I'm going to address this one: $$ x^T A x + v^T \cdot x + f=0 $$

In this, $f$ is proportional to the amount of offset from the axis of the double-cone. That's pretty easy. (The constant of proportion comes from the fact that you can double everything here and still get the same conic, etc.)

The vector $v$ tells you the normal vector to your slicing plane. It's probably a good idea to rescale everything so that the vector $v$ is a unit vector; this removes some of the "proportional" ambiguity from $f$.

What about the matrix $A$? It represents two things: the non-roundness of the cone you're slicing, and the orientation of the resulting conic in the slice plane.

Because $A$ is a symmetric matrix, it can be diagonalized. That means that there are unit vectors $u_1, u_2$ in the plane so that when you form the matrix $U = [u_1, u_2]$ that has those as its columns, the matrix $$ M = U A U^t $$ is diagonal. What this says is that if you use the directions $u_1$ and $u_2$ as your "axes" in the plane, then your conic will have no cross-term -- it'll be "axis aligned", like a parabola whose directrix is horizontal or vertical, or an ellipse whose axes are vertical/horizontal.

So $M$ now has zeros on the off-diagonal entries, and the two diagonal entries might be zero or nonzero. If they're both 0, then your original cone has cone-angle zero, and a slice of it is either a single line or is empty.

If one is nonzero, ... uhh... I'm not sure, and I don't have time to write a lot more, so I'm skipping this case.

If both are positive or both negative, you get a cone that (broadly) looks like $z^2 = x^2 + y^2$; if one's positive and one negative, you get a cone that looks like $z^2 = x^2 - y^2$, which is just $x^2 = y^2 + z^2$, i.e., it's the former case, but with the axes swapped around a bit.

Back to $u_1, u_2$: if you're willing to let these NOT be unit vectors, i.e., to say "I'm going to scale the world up or down along these two perpendicular directions", then $A$ ends up with $0, \pm 1$ on the diagonal, and saying "broadly looks like" becomes "is exactly".

When you do that, it's a good idea to express the vector $v$ in that same coordinate system, i.e., you replace $x$ with $Up$, to get an equation that looks like $$ p^t U^t A U p + v^t \cdot Up + f = 0 $$ which becomes $$ p^t M p + (v^t U^t) \cdot p + f = 0 $$ which becomes $$ p^t M p + w^t \cdot p + f = 0 $$ where $w = Uv$.

So by a change of coordinates, we're looking at slices of the standard double-cone by some plane whose normal vector is $w$.

Does this all help even slightly?

  • $\begingroup$ Thanks for the reply. A few follow up things: 1, You say $f$ is the off-set from the axis of the double cone. Do you mean the distance from the centre of my resulting conic section to $(0,0)$? 2, Secondly, thanks for clarifying that $\vec{v}^T \cdot x$ is the plane part but then $\vec{x}^T A \vec{x}$ should described the cone, shouldn't it? And yet this quadratic form can look like $ax^2+bxy+cy^2$ which will not look like the standard double cone $x^2+y^2=z^2$ which is the thing we want to slice using the plane??? $\endgroup$ – user11128 Dec 17 '16 at 16:39
  • $\begingroup$ And also, if the cone has $b \neq 0$, then it is kind of on it's side according to WolframAlpha meaning that to return it to the standard double cone as you suggest would require a 3-dimensional rotation. Yet, the rotation that they discuss in the literature is just a 2d rotation (x,y)->(x',y'). Can you help me understand how to visualise this? Thank you! $\endgroup$ – user11128 Dec 17 '16 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.