Does a right circular cone only consists of pair of straight lines, hyperbolas, parabola, circles and ellipses? [closed]

I was reading about the conic sections and that a conic section includes pair of straight lines, ellipses, hyperbola, circles and parabola. Are all these 5 components enough to form a right circular cone or are these just the parts of right circular cone? What I mean is : if I put together infinite number of circles, parabolas, hyperbolas,, ellipses and circles, will I get a right circular cone?

closed as unclear what you're asking by Yves Daoust, Winther, Yagna Patel, yoknapatawpha, user228113 Jan 28 '16 at 0:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Do you mean an infinite number of ? – Yves Daoust Jan 27 '16 at 7:39
• "if I put together infinite circles, parabolas, hyperbolas,, ellipses and circles, will I get a right circular cone?" I wonder what you mean by this, it's unclear. Please rephrase. – Yves Daoust Jan 27 '16 at 7:45
• "Put together" is a poorly defined phrase. But you don't need all those things. You can do it with an infinite number of points if they are the right points in the right places. Or just circles if you stack the right ones in the right places. Or hyperbolas if you line them up next to each other just right. Etc. Oh, and if I took a knife and cut a cone like a jackolantern I'd get something other than what you listed. It just wouldn't lie in a plane. – fleablood Jan 27 '16 at 8:15
• It seems to me that "put together" is a perfectly reasonable synonym for "form the union of". It may not be official mathematical jargon, but that's fine with me. – bubba Jan 27 '16 at 11:57
• "You can do it with an infinite number of points if they are the right points in the right places" @fleablood how can we make a 3-dimensional cone by using a one-dimensional entity such as a point?? – Freelancer Feb 4 '16 at 11:24

If you intersect a cone with a plane, the intersection will be one of the following: a parabola, a circle, an ellipse, a hyperbola, a pair of lines (the plane must lie along the axis of the cone), a single line (plane is tangent to cone), or a single unique point (the plane must be perpendicular to the axis, passing through the center).

There are infinitely many parabolas, circles, ellipses, hyperbolas, and lines that can be found in this way. You're asking if we can build a cone using these "parts." Well, intuitively, yes. After all, we just found these parts on the surface! But let's be more specific. Will any arrangement in space of, say, circles generate a cone? No. A random arrangement of many circles would probably look like a scattered mess. Or maybe not. If you're really lucky, your random arrangement could perhaps be a torus: Okay, so what if we're not randomly placing our parts. What if we're carefully placing circles against each other to generate our cone? Then yes, you can build a cone. That is, if you consider the point at the center a circle of radius zero. Using ellipses would work similarly.

What if you use hyperbolas? Sure, that works. Though as you move from one side to the other, the center-most piece will degenerate to two lines. Using parabolas would similarly have a center-most piece that degenerates into a single line.

If you take a single straight line and rotate it about an axis through a point on that line, you'll also map out a cone. In this case, there will be no degeneracy.

There is also a single straight line and a point - both degenerate in the sense that the degree of the equation is less than $2$.

You can consider the cone as cut by a family of parallel planes, and the conics on each plane will combine to give the whole cone. They will all be of the same type, with the plane through the apex giving a degenerate form.

You get a family of pairs of lines by considering cuts by the family of planes which contain the axis of the cone, and in this case all the "conics" are the same.

You can also consider a family of conics generated by the planes which contain a random line in space. In this case you will likely get all conics except a pair of lines (this only when the line goes through the apex, and consider a plane containing your line and the axis of the cone). You might miss a circle in cases where the plane through the apex is perpendicular to the axis and you get a point as a degenerate circle. Otherwise the apex gives you a degenerate ellipse.

• I am not sure but does it mean that in most of the cases we do get a cone? – bluebellae Jan 27 '16 at 7:43
• @bluebellae You start with the cone and cut it into pieces, rather than starting with the pieces and building them up. You automatically get a cone that way, since the various family of planes include every point in space. – Mark Bennet Jan 27 '16 at 8:17

A cone is a surface and as such cannot consist of a finite number of curves.

A conic section is the intersection of a cone with a plane, so it is a planar curve. Depending on the relative positions of the cone and the plane, the section can be a single or a double line, an ellipse, a parabola or an hyperbola (the circle is just a particular case of the ellipse).

A way to reconstruct a cone is to sweep a plane over the whole space (for instance by translation or rotation), a put together all the sections so obtained.

Putting together random lines, ellipses... in any number will just result in a mess.

Yes, a right circular cone consists only of pair of straight lines, hyperbolas, parabola, circles and ellipses when in each case it is cut transverse planes.

Take a cone of a fixed semi-vertical angle $\alpha$ whose symmetry axis is the x-axis

Case 1

Cut it by a plane of variable inclination $\beta$ and constant $C$.

$$y = \tan \beta \,x + C \tag{1}$$

The conic section so formed has the eccentricity

$$\epsilon = \dfrac{\sin\alpha }{\sin \beta} \tag{2}$$

CASE $c \ne 0$ i.e., cutting plane avoids the cone apex:

If $\beta > \alpha$ you get an ellipse.

If $\beta < \alpha$ you get a hyperbola in two disjointed parts.

If $\beta = \alpha$ you get a parabola.

If you rotate any of the three above conics about the x-axis you get back the cone cut segment.

CASE $c = 0$ i.e., cutting plane includes the cone apex:

A pair of straight lines which include an angle $2 \gamma$

$$\tan \gamma = \tan \alpha \,\cos \beta \tag{3}$$

Case 2

You can keep $\beta$ constant and $C$ variable, you get back the cone of stacked conic sections

If you dilate and shift any conic by multiplication factors $(\sin \alpha, \cos \alpha )$ respectively along x-axis

This response would be more convincing with images.

Using expression (4) uploading a (poor quality) image of a two dependent parameter $r,\theta$ cone surface where one of $\beta, C$ is varied at a time, keeping the other fixed. Ellipses are stacked to make up the cone.All type three types of conics are seen on one cone sheet.

(Cone apex is excluded in the first plot. The parabola is not plotted in the second .. due to limitations in expression when some quantities go infinite). Parametric Equations of all possible conics

$$(x,y,z) = (x, \tan \beta\, x - c , \pm \sqrt {x^2( \tan{^2} \alpha -\tan{^2} \beta) + 2 \tan \beta\, c \,x - c^2} ) \tag{4}$$

The process by which you get back the cone from conic sections is in two ways.

Rotation through x-axis rotation, and,

Dilation/Translation keeping conic on x-axis.