Cones

The Elliptic Cone

Elliptic cone

An elliptic cone is the surface formed by the straight lines joining the vertex to each point of an ellipse taken as the base.

The Standard Equation

The standard equation of the elliptic cone

x2a2+y2b2−z2c2=0\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0

where a,b,c>0a, b, c > 0.

The Geometric Properties

  • The vertex: at the origin (0,0,0)(0, 0, 0)
  • The generators: all the straight lines through the vertex
  • Symmetry: symmetric about the three coordinate planes
  • The direction it opens: it opens in both the positive and the negative zz-directions

The Properties of Cones

The Nature of the Sections

The sections of a cone

The sections of a cone by planes:

  • by planes parallel to the xyxy-plane: ellipses
  • by planes parallel to the xzxz-plane: hyperbolas
  • by planes parallel to the yzyz-plane: hyperbolas
  • by planes through the vertex: two intersecting straight lines

The Behaviour at Infinity

A cone has the behaviour at infinity: as a point moves away from the vertex, the surface approaches infinity.

The Equation of a Generator

Every generator of the cone satisfies:

xa=yb=zc\frac{x}{a} = \frac{y}{b} = \frac{z}{c}

Special Cases

The Circular Cone

When a=ba = b in the elliptic cone, it becomes a circular cone:

x2a2+y2a2−z2c2=0\frac{x^2}{a^2} + \frac{y^2}{a^2} - \frac{z^2}{c^2} = 0

that is: x2+y2=a2c2z2x^2 + y^2 = \frac{a^2}{c^2} z^2

The Equilateral Cone

When a=b=ca = b = c:

x2+y2−z2=0x^2 + y^2 - z^2 = 0

This is an equilateral elliptic cone.

Applications of Cones

Geometric Construction

In three-dimensional modelling, cones are used to create cone-shaped objects.

Optics

In optics, cones are used to describe the propagation paths of light rays.

Engineering Applications

In mechanical engineering, cones are used to design conical parts.


Exercises

Exercise 1

Determine the type of the surface x2+y2−z2=0x^2 + y^2 - z^2 = 0.

conerecognising the equationthe equilateral cone

Approach:

Recognise the standard equation form of the cone.

Steps:

  1. The equation: x2+y2−z2=0x^2 + y^2 - z^2 = 0
  2. It can be written as: x212+y212−z212=0\frac{x^2}{1^2} + \frac{y^2}{1^2} - \frac{z^2}{1^2} = 0
  3. This is the standard equation of the equilateral elliptic cone

Answer: this is an equilateral elliptic cone

Exercise 2

Find an equation of one generator of the elliptic cone x24+y29−z216=0\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{16} = 0.

conegeneratorsparametric equations

Approach:

Use the parametric form of the equation of a generator.

Steps:

  1. The equation of the generator: x2=y3=z4\frac{x}{2} = \frac{y}{3} = \frac{z}{4}
  2. Introducing the parameter tt, we get: x=2tx = 2t, y=3ty = 3t, z=4tz = 4t
  3. This is one generator of the cone

Answer: the equation of the generator is x2=y3=z4\frac{x}{2} = \frac{y}{3} = \frac{z}{4}


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
x2a2+y2b2−z2c2=0\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 0Elliptic coneelliptic coneThe standard equation of the elliptic cone
xa=yb=zc\frac{x}{a} = \frac{y}{b} = \frac{z}{c}Generator equationgenerator equationThe parametric equation of a generator of the cone

Found an issue with this content?

Tell us what could be improved.