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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.

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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?

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  • $\begingroup$ 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

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