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Styding such curves as ellipse, parabola, hyperbola I am a bit confused of what their general equation is called, and mainly - where it comes from:

$$Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey +F = 0 \space (1)$$

(Also, in English sources, monomials may do not have 2 coefficient)

First, in Russian Wikipedia it called second order curve equation (maybe degree, instead of order), but when I try to locate English variant of that article, it redirects me to Conic section article, so I am a bit confused what I am asking about, how it should be properly called. Perhaps, second order curve is generalization of conic section equation?

Next, my basic question is where equation $(1)$ is derived from? If $(1)$ is a conic section, then, probably, it is a solution of mutual points - intersection of a plane and a cone.

I think, it comes from something like this:

$$\dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} - \dfrac{z^2}{C^2} = 0 \space Cone$$ $$Dx + Ey + Fz + G = 0 \space Plane$$ $$\dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} - \dfrac{z^2}{C^2} = Dx + Ey + Fz + G \space (2) \space Cone \space and \space plane \space mutual \space points$$

Unfortunately, first - plot of $(2)$ with all coefficients equal 1, i.e. $x^2 + y^2 - z^2 = x+y+z+1$ does not look like a conic section in 3D, second - conic section equation is 2D-equation, it does not have third variable, so $(2)$ anyway is not acceptable.

Considering problem, mentioned in previous paragraph, conic section seems to be not just mutual points of plane and cone, but mutual point of plane and cone in plane "coordinates system", i.e. relatively to plane.

So, from where do I get $(1)$ equation? Why it exactly like this? From $Ax^2 + 2Bxy + Cy^2$, it seems, there was something like $(x+y)^2$, but then it somehow was multiplied separately by constants $A$, $B$ and $C$.

sahgasdvsadgv
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  • What you call the coefficients does not matter. There are many forms, just as there are many forms to parabolas or line equations. It all depends on what information you want to encode in the coefficients. – John Alexiou Dec 27 '23 at 14:08
  • Why slicing a cone gives an ellipse – John Alexiou Dec 27 '23 at 14:10
  • They are usually called conics and their multidimensional counterparts are called quadrics, they are all second degree curves/variëties since they are defined by a second degree polynomial. – Vincent Batens Dec 27 '23 at 14:20
  • @JohnAlexiou, 1) As for Your first comment - my question is why that equation is what it is? Why it is exactly that, where it derived from? Or someone just stated "let it be the next way"? 2) 12 minutes, I probably will not watch it, because of title - I have already successfully derived ellipse equation as a set of points, which sum of distances to two points, called focuses are constant - from $\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2a$. Same for hyperbola. I also understand intuitively why one of conic section is an ellipse, so, probably, I will not find anything, related to the question. – sahgasdvsadgv Dec 27 '23 at 14:34
  • @JohnAlexiou, as I read, all conic sections equations are derived from equation $(1)$ from my post. More accurate - $(1)$ is generalized form of all conic sections. I am asking, where $(1)$ is derived from, not from where some particular cases, such an ellipse are derived from. – sahgasdvsadgv Dec 27 '23 at 14:38
  • @VincentBatens, who "they"? – sahgasdvsadgv Dec 27 '23 at 14:39
  • The left hand side of your equation (1) is the most general form of a quadratic in two variables. It isn't derived from anything. Putting it equal to 0 is a curve in 2D. Depending on the coefficients this can be reduced using translations and rotations to a hyperbola, parabola, ellipse or degenerates into two straight lines. – Paul Dec 27 '23 at 14:40
  • @Paul 1) Seems like true. Is it school algebra course? From quadratic equations, I remember only $ax^2 + bx + c = 0$ and Newtons binomials – sahgasdvsadgv Dec 27 '23 at 14:57
  • @Paul 2) Second order surface, as I understand also is not derived from somewhere? – sahgasdvsadgv Dec 27 '23 at 14:59
  • @Paul 3) However, can $(1)$ somehow be obtained as solution of intersection of plane and cone? By definition, it should – sahgasdvsadgv Dec 27 '23 at 15:00
  • A quadratic equation would be a special case with $ax^2+bx-y+c=0$. You are missing $y$ in your first comment. There is also no $z = ...$ so it is not a surface. – Paul Dec 27 '23 at 15:02
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    @JohnAlexiou, I watched, proof with spheres is interesting of course (although I did not understand it completely, because of language, need to watch more), however question was not about that – sahgasdvsadgv Dec 27 '23 at 15:03
  • @Paul, I did not miss, see Quadratic equation. It is an equation, which roots are $x=\frac{-b±\sqrt{b^2-4ac}}{2a}$ – sahgasdvsadgv Dec 27 '23 at 15:06
  • $ax^2+bx+c=0$ defines two or possibly one (complex) point, you have the formula for those point(s). On the other hand, $y = ax^2+bx+c$ defines a curve of points $(x,y)$. Similarly, your form (1) relates x and y so that it is also a curve of points. If you say $F(x)=0$ for a single variable x that defines a set of points, the roots of F. If you say $F(x,y)=0$ that defines a relationship between x and y, a curve in 2D. If you say $z=F(x,y)$ that defines a surface of points, one z value for each pair $(x,y)$. – Paul Dec 27 '23 at 15:16
  • Note that the xy variables are planar coordinates and the xyz variables are spatial coordinates. You need to distinguish between the two, by using different names for different things. The same applies to the coefficients ABC.. – John Alexiou Dec 27 '23 at 15:43
  • @sahgasdvsadgv "they" referred to collections of points that satisfy equations like you wrote in your post. – Vincent Batens Dec 28 '23 at 19:34
  • The second-degree polynomial curves are typically called conics in English because they (almost) all represent the intersection of a right circular cone in $x,y,z$ coordinates with the plane $z = 0$. (There are a few special cases such as two parallel lines that are intersections with a "cone" only in a degenerate sense.) See Confusion about the conic equation. – David K Dec 29 '23 at 21:16

1 Answers1

1

The planar equation of a conic section comes from the quadratic form of the equation of a cone (in terms of $x^2$, $y^2$, and $z^2$ and all lower order combinations) and the equation of the plane which can solve for one variable in terms of the other two (for example $z = f(x,y)$).

The result in the equation of the cone is constrained to a plane in terms of two variables (like $x^2$ and $y^2$ and lower terms) only.

Given a 3D point $\boldsymbol{r}$, the equation of a cone with apex $\boldsymbol{a}$, unit direction vector $\boldsymbol{\hat g}$ and cone angle $\theta$ is

$$ \array{\text{cone: } & (\boldsymbol{r} - \boldsymbol{a})\cdot\boldsymbol{\hat g} = \| \boldsymbol{r} - \boldsymbol{a} \|\cos \tfrac{ \theta}{2} } \tag{1} $$

and the equation of a plane with unit normal direction $\boldsymbol{\hat n}$ and offset from origin $d$ is

$$ \array{\text{plane: } & \boldsymbol{r} \cdot \boldsymbol{\hat n} = d } \tag{2} $$

If you squared (1) you will transform it into its quadratic form

$$\left( (\boldsymbol{r} - \boldsymbol{a})\cdot\boldsymbol{\hat g} \right)^2 = (\boldsymbol{r} - \boldsymbol{a})\cdot(\boldsymbol{r} - \boldsymbol{a})\cos^2 \tfrac{ \theta}{2} \tag{3}$$

which contains terms with $x^2$, $y^2$, $z^2$, $x y$, $y z$, $z x$, $x$, $y$, $z$, and constants.

and if (2) is used to eliminate one of the coordinates, for example

$$z = \frac{d - n_x\; x - n_y\; y}{n_z}$$

then (3) is going to become a planar conic section in terms of $x^2$, $y^2$, $x y$, $x$, $y$, and constants.

Example

A cone along the z-axis intersects a plane as shown below

fig1

The dimensions are shown below

fig2

The equation of the cone is given by the dot product cosine rule

$$\begin{pmatrix}0\\ 0\\ 1 \end{pmatrix}\cdot\begin{pmatrix}x\\ y\\ z \end{pmatrix}=\left(\sqrt{x^{2}+y^{2}+z^{2}}\right)\cos\left(30°\right)$$

which when both sides are squared gives

$$ z^2 = (x^2+y^2+z^2) \tfrac{3}{4} $$

And the equation for the plane is

$$ \begin{pmatrix}\text{-}\sin45°\\ 0\\ \cos45° \end{pmatrix}\cdot\begin{pmatrix}x\\ y\\ z \end{pmatrix}=5 $$

which is simplified to

$$ \tfrac{\sqrt{2}}{2} (z-x) = 5 $$

and then

$$ z = x + 5 \sqrt{2} $$

which is used in the equation for the cone to produce

$$ ( x + 5 \sqrt{2} )^2 = \left(x^2+y^2+(x+5\sqrt{2})^2\right) \tfrac{3}{4}$$

and when all the terms are brought to one side

$$ \frac{x^2}{2} - \frac{5 \sqrt{2} x}{2} + \frac{3 y^2}{4} - \frac{25}{2} = 0 $$

which looks like an ellipse. But it is defined in the cartesian xyz space, and not on the 2D planar coordinates.

I found that the planar coordinates $(u,v)$ can be transformed to the cartezian space with

$$\begin{pmatrix}x\\ y\\ z \end{pmatrix}=\begin{pmatrix}u+\frac{5\sqrt{2}}{2}\\ v\\ u+\frac{15\sqrt{2}}{2} \end{pmatrix}$$

such that $(u,v)=(0,0)$ represents the center of the ellipse. The above obeys the plane equation and transforms the ellipse into

$$ \frac{u^2}{2} + \frac{3 v^2}{4} - \frac{75}{4} = 0 $$

The above is brought into the standard ellipse form of

$$ \left( \frac{u}{a} \right)^2 + \left( \frac{v}{b} \right)^2 =1$$ with the parameters $a = \frac{5 \sqrt{6}}{2}$ and $b = 5$.

John Alexiou
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  • Can You please supplement Your answer with solution in coordinates form instead of vector? – sahgasdvsadgv Dec 29 '23 at 12:07
  • @sahgasdvsadgv - the expressions become rather large and cumbersome in component form. I deliberately chose to show an overview of the process using the vector forms to explain why a conic section is a planar curve with quadratic terms only. – John Alexiou Dec 29 '23 at 13:33
  • I need some time to make decision to accept the post – sahgasdvsadgv Jan 01 '24 at 15:52