Determine if an equation describes a conic section

[Happiness]

Given an equation, we can determine which type of conic section it describes by calculating $d=B^2-4AC$ (see attachment). However, the theorem demands that the equation describes a conic section. So how do we show that it does?

Does there exist a counterexample where $B^2-4AC<0$, but it is neither a circle or an ellipse?

EDIT: I found that if $d<0$ there is no need to check that a conic section exists because $d<0$ implies a conic section exists.

But it remains to find out whether there is a need to check if a conic section exists when $d=0$ or $d>0$, given that $B\neq0$.

 

[Simon Bridge]

Well done.

 

[Happiness]

Well done.

But it remains to find out whether there is a need to check if a conic section exists when $d=0$ or $d>0$, given that $B\neq0$.

 

[micromass]

Uh, well, it is always possible that the conic section is degenerate. For example $x^2 + y^2 = -1$ is a conic section that should qualify as an ellipse, but there are no points satisfying this equation. Somewhat similar with $x^2 + y^2 = 0$, where there is only one point. These are called degenerate ellipses. This situation is completely resolved by complex numbers however.

 

[micromass]

But it remains to find out whether there is a need to check if a conic section exists when $d=0$ or $d>0$, given that $b\neq0$.

I really don't get what you mean with "a conic existing". If you have an equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0$, then you have your conic. So what's all this things about existing.

 

[Happiness]

I really don't get what you mean with "a conic existing". If you have an equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0$, then you have your conic. So what's all this things about existing.

Without a proof of existence, it may be possible that the equation describe a curve that is neither a hyperbola nor a parabola nor an ellipse nor a circle nor a pair of intersecting lines; it could be a curve that looks like a hyperbola or a parabola but is neither or it could be an oval that is not an ellipse.

 

[micromass]

Without a proof of existence, it may be possible that the equation describe a curve that is neither a hyperbola nor a parabola nor an ellipse nor a circle nor a pair of intersecting lines; it could be a curve that looks like a hyperbola or a parabola but is neither or an oval that is not an ellipse.

Ok, so you want to prove that the equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0$ always describes either an ellipse, hyperbola or parabola. And those can either be degenerate (pairs of lines) or nondegenerate. Good. But then you first need to tell us what your definition of ellipse/hyperbola/parabola is.

 

[Happiness]

Ok, so you want to prove that the equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0$ always describes either an ellipse, hyperbola or parabola. And those can either be degenerate (pairs of lines) or nondegenerate. Good. But then you first need to tell us what your definition of ellipse/hyperbola/parabola is.

They are defined by their standard representations:

from which we could generalise the equations by replacing $x$ by $x'\cos\theta-y'\sin\theta$ and replacing $y$ by $x'\sin\theta+y'\cos\theta$, which corresponds to a rotation of the OXY coordinate system by $\theta$ clockwise to give the OX'Y' system. It remains to show that the equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0$ can always be written as one of these generalised equations.

Equivalently, we could define them geometrically: they are the loci of a point whose distance from a given point (the focus) is a fraction $e$ (eccentricity) of its distance from a line (the directrix).

 

[micromass]

They are defined by their standard representations:
View attachment 100953
from which we could generalise the equations by replacing $x$ by $x'\cos\theta-y'\sin\theta$ and replacing $y$ by $x'\sin\theta+y'\cos\theta$, which corresponds to a rotation of the OXY coordinate system by $\theta$ clockwise to give the OX'Y' system.

Equivalently, we could define them geometrically: they are the loci of a point whose distance from a given point (the focus) is a fraction $e$ (eccentricity) of its distance from a line (the directrix).

OK nice. And indeed, we can bring back any equation of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey +F=0$ to one of those standard representations. Have you see matrix diagonalization??

 

[Happiness]

Have you see matrix diagonalization??

Yes, but I'm afraid I'm not well versed in it.

 

[micromass]

Yes, but I'm afraid I'm not well versed in it.

OK, let me try to explain it then. First, note that any equation of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ can be written as

$$\left(\begin{array}{cc} x & y\end{array}\right)\left(\begin{array}{cc} A & B/2\\ B/2 & C\end{array}\right)\left(\begin{array}{c} x\\ y\end{array}\right) +\left(\begin{array}{cc} D & E\end{array}\right)\left(\begin{array}{c} x\\ y\end{array}\right) + F = 0$$

Using diagonalization, we can diagonalize the matrix $\delta = \left(\begin{array}{cc} A & B/2\\ B/2 & C\end{array}\right)$. Note that this is a symmetric matrix, so we can write $\delta = P^{-1}\delta^\prime P$, where $P$ are rotation/reflection matrices and $\delta^\prime$ is a diagonal matrix. So by the coordinate change $(x,y)\rightarrow P(x,y)$, we can write the above equation with $\delta$ changed by a diagonal matrix. This comes down to eliminating the $xy$ term. Thus we get

$$A' x^2 + C' y^2 + D' x + E'y + F' = 0$$

Now we complete the squares to obtain the standard form.

I think an example will be much more useful than this general technique. Take

$$3x^2 - 10xy + 3y^2 + 14x - 2y + 3 = 0.$$

Writing this in matrix form, we get
$$\left(\begin{array}{cc} x & y\end{array}\right)\left(\begin{array}{cc} 3 & -5\\ -5 & 3\end{array}\right)\left(\begin{array}{c} x\\ y\end{array}\right) +\left(\begin{array}{cc} 14 & -2\end{array}\right)\left(\begin{array}{c} x\\ y\end{array}\right) + F = 0$$

The usual techniques of diagonalization give us

$$\left(\begin{array}{cc} 3 & -5\\ -5 & 3\end{array}\right) = \left(\begin{array}{cc} 1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2}\end{array}\right)\left(\begin{array}{cc} 8 & 0\\ 0 & -2\end{array}\right)\left(\begin{array}{cc} 1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2}\end{array}\right)^{-1}$$

Note that in fact, this is

$$\left(\begin{array}{cc} 3 & -5\\ -5 & 3\end{array}\right) = \left(\begin{array}{cc} \cos(-\pi/4) & -\sin(-\pi/4) \\ \sin(-\pi/4) & \cos(-\pi/4)\end{array}\right)\left(\begin{array}{cc} 8 & 0\\ 0 & -2\end{array}\right)\left(\begin{array}{cc} \cos(-\pi/4) & -\sin(-\pi/4) \\ \sin(-\pi/4) & \cos(-\pi/4)\end{array}\right)^{-1}$$

So by rotating our axis $-\pi/4$ radians, which comes down to changing variables

$$\left(\begin{array}{c} x\\ y \end{array}\right)=\left(\begin{array}{cc} \cos(-\pi/4) & -\sin(-\pi/4) \\ \sin(-\pi/4) & \cos(-\pi/4)\end{array}\right)\left(\begin{array}{cc} x'\\ 'y\end{array}\right)$$

Then we have that

$$\left(\begin{array}{cc} x & y\end{array}\right)\left(\begin{array}{cc} 3 & -5\\ -5 & 3\end{array}\right)\left(\begin{array}{c} x\\ y\end{array}\right) +\left(\begin{array}{cc} 14 & -2\end{array}\right)\left(\begin{array}{c} x\\ y\end{array}\right) + F = 0$$

Is equivalent to

$$\left(\begin{array}{cc} x' & y'\end{array}\right)\left(\begin{array}{cc} \cos(-\pi/4) & -\sin(-\pi/4) \\ \sin(-\pi/4) & \cos(-\pi/4)\end{array}\right)^T\left(\begin{array}{cc} 3 & -5\\ -5 & 3\end{array}\right)\left(\begin{array}{cc} \cos(-\pi/4) & -\sin(-\pi/4) \\ \sin(-\pi/4) & \cos(-\pi/4)\end{array}\right)\left(\begin{array}{c} x'\\ y'\end{array}\right) +\left(\begin{array}{cc} 14 & -2\end{array}\right)\left(\begin{array}{cc} \cos(-\pi/4) & -\sin(-\pi/4) \\ \sin(-\pi/4) & \cos(-\pi/4)\end{array}\right)\left(\begin{array}{c} x'\\ y'\end{array}\right) + F = 0$$

or

$$\left(\begin{array}{cc} x' & y'\end{array}\right)\left(\begin{array}{cc} 8 & 0\\ 0 & -2\end{array}\right)\left(\begin{array}{c} x'\\ y'\end{array}\right) +\left(\begin{array}{cc} 14 & -2\end{array}\right)\left(\begin{array}{cc} \cos(-\pi/4) & -\sin(-\pi/4) \\ \sin(-\pi/4) & \cos(-\pi/4)\end{array}\right)\left(\begin{array}{c} x'\\ y'\end{array}\right) + F = 0$$

Eliminating the matrices again (and writing $x,y$ instead of $x', y'$ for convenience), we get

$$8x^2 -2y^2 +8\sqrt{2}x+6\sqrt{2}y +3 = 0.$$

Completing the squares, we get

$$8(x + 1/\sqrt{2})^2 - 2(y -3/\sqrt{2})^2 =-8.$$

By a simple translation, that is changing $(x + 1/\sqrt{2}, y - 3\sqrt{2})$ by $(x,y)$, we get

$$8x^2 - 2y^2 = -8.$$

This is a hyperbola

 

[Happiness]

It seems like I've found a counterexample: $2x^2+12xy+18y^2+14x+42y+20=0$ describes a pair of parallel lines, which is not a conic section. Or could this also be resolved by using complex numbers? If so, is it classified as a "complex" ellipse or parabola or hyperbola or a pair of "complex" intersecting lines?

 

[member 587159]

It seems like I've found a counterexample: $2x^2+12xy+18y^2+14x+42y+20=0$ describes a pair of parallel lines, which is not a conic section. Or could this also be resolved by using complex numbers? If so, is it classified as a "complex" ellipse or parabola or hyperbola or a pair of "complex" intersecting lines?

Parallel lines are a conic section. It's a special parabola.

 

[member 587159]

Do you know the matrix that represents a conic section? If so, if the determinant of this matrix is zero, then the equation of the conic section represents 2 lines.

 

[micromass]

It seems like I've found a counterexample: $2x^2+12xy+18y^2+14x+42y+20=0$ describes a pair of parallel lines, which is not a conic section. Or could this also be resolved by using complex numbers? If so, is it classified as a "complex" ellipse or parabola or hyperbola or a pair of "complex" intersecting lines?

So you have three types of nondegenerate conics:
- Ellipses
- Hyperbolas
- Parabolas

Then you have three types of degenerate conics:
- Ellipses: these can either consist of a single point or be completely empty
- Hyperbolas: these are a pair of intersecting lines. For example $x^2 - y^2 = 0$.
- Parabolas: these are either a pair of parallel lines, for example $x^2 - x =0$. Or just one line, for example $x^2 = 0$. Or empty, for example $x^2 = -1$.

Checking whether the conic $Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0$ is degenerate can be done by computing a determinant, as Math_QED explained. In particular, the conic is degenerate if and only if
$$\text{det}\left(\begin{array}{ccc} A & B/2 & D/2\\ B/2 & C & E/2\\ D/2 & E/2 & F\end{array}\right)=0.$$

 

[Simon Bridge]

This has been interesting ... I've been thinking some more about the original question in post #1:
In the graphic/attachment, there is a section from some notes or something where the author keeps saying "if the conic exists".
I suspect this is the source of the confusion.

Normally, if it can be written in the standard form for a conic section, then it is a conic section: pretty much by definition.
Since the author starts out by assuming the standard form, it is unclear what they mean by "if it exists". It must exist: you already have the standard form.

The trouble here, I suspect, stemms from the way mathematicians treat definitions. I'll illustrate:
It has been observed that the "standard form" allows parallel lines to be a conic section ... this is a case in point. If you define a conic section as the intersection of a plane and a cone then it won't work for a regular cone that you can hold in your hand... the object being cut here is a cylinder. Most people would not think of a cylinder as a type of cone.

Most people would agree, or at least accept, that a "cone" is the solid of rotation formed by a sloping line.
If we define the line as passing through points P and R and make them, say, P=(0,0,p) and R=(r,0,0) and make the solid of rotation about the z axis ... you can see how that is a cone.
The intersection of the cone with the x-y plane is a circle radius r.
The intersection of a vertical plane (any plane perpendicular to the x-y plane) with the cone is a hyperbola - if we use that as a definition, then the crossed lines conic must be a special case of a hyperbola.

Imagine changing the cone by changing the value of p (sliding P up and down the z-axis) but keeping r fixed.
... as p gets small, the cone flattens out into a plane.
... as p gets very large, the bit of the cone by the x-y plane becomes a cylinder radius r.

The intersection of a vertical plane with a cylinder is a single line or a pair of parallel lines.
So this must be a special case of a hyperbola.

What I am trying to illustrate here is that mathematical definitions of things can include objects that common sense would normally exclude.

What does this mean for post #1?
It translates your question into: what does the author mean by "if it exists"?

The answer is that we cannot really know: not enough information.
It could just be overspecifying the situation - i.e. it means "if you can write it in the standard form above"
If could mean that degenerate cases are not being included here.
Or it could means something else.

Usually the author of a textbook will make a statement about existence earlier in the text, passages like this are referring back.
Where did you get the excerpt from?

 

[Happiness]

Where did you get the excerpt from?

http://hotmath.com/hotmath_help/topics/conic-sections-and-standard-forms-of-equations.html

 

[micromass]

Wow, many errors on that page. Saying a circle is a different type than an ellipse is particularly inexcusable.

 

[Simon Bridge]

Wow, many errors on that page. Saying a circle is a different type than an ellipse is particularly inexcusable.

I didn't see that bit - but it is kinda implied, and that is the usual way conic sections is usually taught.

@Happiness
I can't tell from that what the author means by "if [the conic] exists"... probably just a bit of nonsense.
I'd suggest trying a number of other sources ... comparing sources is a big part of science literacy anyway.