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Let me start with some basic definitions:

Definition 1. A conic section is the curve resulting from the intersection of a plane and a cone.

Definition 2. A conic section is the set of all points in a plane with the same eccentricity with respect to a particular focus and directrix.

Definition 3. A conic section is the set of points $(x,y)$ satisfying the implicit formula

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

All these definitions are familiar to me and they are taken from a paper of Mzuri S. Handlin, Conic Sections Beyond $\mathbb{R}^2$.

Now, in by university textbook I have the following theorem,

Theorem. Any conic has an equation of the form

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

where $A,B,C,D,E$ and $F$ are real numbers and $A,B$ and $C$ are not all zero. Conversely, the set of all points in $\mathbb{R}^2$ whose coordinates $(x,y)$ satisfy an equation of the form that above, is a conic.

Proof to this theorem is omitted, and reference are not included. I've been searching for a proof, but failed to find one.

It seems that the above theorem make motivate us to define definition 3, right?

If one can provide me a reference for the proof of the theorem or sketch guidelines for possible proof?

Thank you.

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  • $\begingroup$ How does your textbook define a conic ? $\endgroup$ – Yves Daoust Jan 13 '17 at 14:38
  • $\begingroup$ @YvesDaoust: The textbook start straight away with conic section, a non formal definition given: Conic section is the collective name given to the shapes that we obtain by taking different plane slices through a double cone. $\endgroup$ – Salech Rubenstein Jan 13 '17 at 14:46
  • $\begingroup$ This is your definition 1, isn't it ? $\endgroup$ – Yves Daoust Jan 13 '17 at 14:50
  • $\begingroup$ Yes, that is true. $\endgroup$ – Salech Rubenstein Jan 13 '17 at 14:51
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    $\begingroup$ Then write the equation of a cone in general position and set $z=0$. $\endgroup$ – Yves Daoust Jan 13 '17 at 14:52
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In my opinion, the theorem as stated fails to account adequately for some special cases. But we'll deal with those as we proceed.

In order to avoid having to write "ellipse or circle" in various places, I'll use the convention that a circle is a special case of an ellipse.


Part 1. We wish to show that any conic has an equation of the form $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0.$$

Consider a right circular cone of some arbitrary vertex angle in general position, and let $\gamma$ be the intersection of the cone with the $x,y$ plane. Perform an isometry $M$ in $x,y,z$ space that translates the vertex of the cone to the origin and rotates the axis of the cone onto the $z$ axis.

The image of the original conic section $\gamma$ under $M$, the curve $M(\gamma),$ is the intersection of some cone of the form $x^2 + y^2 - k^2z^2 = 0$ with some plane of the form $px + qy + rz + s = 0.$

We can perform the inverse isometry $M^{-1}$ to take $M(\gamma)$ back to $\gamma.$ The curve $\gamma$ satisfies the previously given equations in rotated and translated coordinates, that is, it satisfies \begin{gather} x'^2 + y'^2 - k^2z'^2 = 0 \tag1 \\ \text{and} \\ px' + qy' + rz' + s = 0 \tag2 \end{gather} where each of the variables $x',$ $y',$ and $z'$ is a first-degree polynomial over the variables $x,$ $y,$ and $z.$

Substitute the appropriate polynomial over $x,$ $y,$ and $z$ for each of the variables $x',$ $y',$ and $z'$ in the preceding equations of $\gamma$; we will then find that Equation $1$ becomes a new quadratic equation in $x,$ $y,$ and $z$ and Equation $2$ is the equation $z=0.$ Then make the substitution $z=0$ in Equation $1$; the result is something of the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0.$

As for the condition that $A,$ $B,$ and $C$ are not all zero, if $A=B=C=0$ we obtain the equation of a line. But technically, this satisfies Definition 1 from the question: a line is the intersection of a cone with a plane through the vertex of the cone and tangent to a circle that generates the cone. It also satisfies Definition 2 if you consider an infinite eccentricity to be acceptable, as some sources do, or if you place the focus on the directrix.

The objection to the "conic" described by $Dx+Ey+F=0$ seems to be solely that is is not one of the "expected" conics, that is, it is not a non-degenerate circle, ellipse, parabola, or hyperbola. Insisting on only these "expected" conics leads to complications in the second part of the theorem, however.


Part 2. For the converse implication, let at least one of the real numbers $A,$ $B,$ and $C$ be non-zero, and let the real polynomial $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ be the equation of a curve $\gamma.$

It will be convenient to use the fact that an ellipse, parabola, or hyperbola given by any equation of the form $Ax^2+Bxy+Cy^2+H=0$ remains an ellipse, parabola, or hyperbola (respectively) under any invertible linear transformation (that is, if the variables $x$ and $y$ are replaced in that equation by any two independent linear combinations of $x$ and $y$) and under any translation. That's a generally useful fact, but giving a full proof if it would make this answer much longer, so let's just suppose we already know it.

Case 1. Suppose $B^2 - 4AC \neq 0.$ We will see that the coefficients $D$ and $E$ are determined by a translation of the center of the conic away from the origin of the $x,y$ plane, and that there is an inverse translation that eliminates these terms.

Because $B^2 - 4AC \neq 0,$ the matrix $\begin{pmatrix} 2A & B \\ B & 2C \end{pmatrix}$ is invertible, and there exist two real numbers $m$ and $n$ that satisfy the simultaneous linear equations \begin{align} 2Am + Bn &= D, \tag3 \\ Bm + 2Cn &= E. \tag4 \end{align}

Let $T(\gamma)$ be the translation of $\gamma$ by $m$ units in the $x$ direction and $n$ units in the $y$ direction. The equation of $T(\gamma)$ is $$A(x-m)^2+B(x-m)(y-n)+C(y-n)^2+D(x-m)+E(y-n)+F=0.$$ Multiply this out to obtain an equation of the form $Ax^2+Bxy+Cy^2+D'x+E'y+F'=0,$ where \begin{align} D' &= D - 2Am - Bn, \\ E' &= E - Bm - 2Cn, \ \text{and} \\ F' &= Am^2 + Bmn + Cn^2 - Dm - En + F \end{align} But Equations $3$ and $4$ imply that $D'=E'=0,$ so the equation of $T(\gamma)$ is $$ Ax^2+Bxy+Cy^2+F'=0. $$

Now if $A=0$ then the equation of $T(\gamma)$ is $(Bx+Cy)y + F'=0,$ so $T(\gamma)$ is a hyperbola. But if $A\neq0,$ we can "complete the square" as follows. Let $\alpha = \sqrt{\lvert A\rvert}.$ Then $$\left(\alpha x + \frac{B}{2\alpha}y\right)^2 = Ax^2 + Bxy + \frac{B^2}{4A}y^2,$$ and therefore the equation of $T(\gamma)$ is $$ \left(\alpha x + \frac{B}{2\alpha}y\right)^2 + \left(C - \frac{B^2}{4A}\right)y^2 + F' = 0. $$ We can also write this in the form $$ \bar x^2 - \frac{B^2 - 4AC}{4A}y^2 + F' = 0 \tag5$$ where $\bar x = \alpha x + \frac{B}{2\alpha}y$. If $B^2-4AC > 0$ this is the equation of a hyperbola if $F\neq0$ but it is the equation of two lines through the origin if $F=0.$ If $B^2-4AC < 0,$ Equation $5$ is the equation of an ellipse if $F'<0,$ a single point if $F'=0,$ but the empty set if $F'>0.$ As these cases describe the possible shapes of $T(\gamma)$, in the same cases $\gamma$ (which is congruent to $T(\gamma)$) must also be a hyperbola, an ellipse, a point, or the empty set, respectively.

Case 2. Let $B^2 - 4AC = 0.$ Since we have assumed $A,$ $B,$ and $C$ are not all zero, $A$ and $C$ cannot both be zero. Without loss of generality, suppose $A \neq 0.$

Let $\alpha = \sqrt{\lvert A\rvert}$ and let $\beta = \frac{B}{2\alpha}.$ Then $2\alpha\beta = B$ and $\beta^2 = C,$ so the equation of $\gamma$ can be written $$(\alpha x + \beta y)^2 + Dx + Ey + F = 0. \tag6$$ If $\alpha E \neq \beta D$ then Equation $6$ gives $Dx + Ey$ as a quadratic function of $\alpha x + \beta y,$ so it is the equation of a parabola. But if $\alpha E = \beta D$ then Equation $6$ is a quadratic equation in $\alpha x + \beta y$ and it is the equation of either a single straight line, two parallel lines, or the empty set.


So we see that the theorem as stated is not quite right. First of all, there are equations of the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ that have no solutions in real numbers $x$ and $y.$ There are several special cases to consider: \begin{align} P:\quad & A=B=C=0, \\ Q:\quad & B^2 - 4AC > 0 \quad \text{and} \quad F = 0, \\ R:\quad & B^2 - 4AC < 0 \quad \text{and} \quad F \leq \frac{BDE-AE^2-CD^2}{B^2 - 4AC}, \\ S:\quad & B^2 - 4AC = 0 \quad \text{and} \quad AE = CD. \end{align} If any of the propositions $P,$ $Q,$ $R,$ or $S$ is true, it can be shown through algebra that the equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is one of the cases whose solution set has already been shown to be one or two lines, a single point, or the empty set. But if all four propositions are false then we have an ellipse, a parabola, or a hyperbola. The theorem as stated, however, considers only proposition $P,$ so it does not give sufficient conditions to determine whether the equation's solution is a non-degenerate conic (an ellipse, parabola, or hyperbola) or one of the other possible shapes.

If we accept that the conic section might be a degenerate form, however, then all the possible solution sets can be interpreted as conics. This includes the empty set, which is obtained by a cone with axis parallel to the $z$ axis and vertex angle equal to a right angle, and parallel lines, which are obtained by intersecting the plane with a cylinder, which can be considered a cone with vertex at infinity (which it actually is, in projective geometry). A single point, single line, and intersecting lines can all be obtained by placing the vertex of the cone on the $x,y$ plane. Under this interpretation, it is not necessary to exclude the case in which $A=B=C=0.$


The answers to the following question may also be of interest: Confusion with the various forms of the equation of second degree

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  • $\begingroup$ Wonderful explanation, for what it's worth, +1 $\endgroup$ – Kevin Jan 27 '17 at 14:38

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