Unit 5: The Cone and The Cylinder

The Cone

Definition

A **cone** is a 3D surface generated by a moving straight line (called the **generator**) that always passes through a fixed point (the **vertex**) and intersects a given curve (the **guiding curve** or **base**).

Equation of a Cone with Given Vertex and Base

This is a standard problem of finding the locus. We use the method of elimination.

Method:

1. Let the vertex be V(α, β, γ).
2. Let the guiding curve (base) be given by f(x, y) = 0 in the plane z = 0. (Example: x² + y² = a², z = 0).
3. Take any generator line. Since it passes through the vertex, its equations are:
   (x - α)/l = (y - β)/m = (z - γ)/n
4. This generator must intersect the base. Find the point where it intersects the plane z = 0.
   Set z = 0 in the line equations:
   (x - α)/l = (y - β)/m = (0 - γ)/n
   Solve for x and y:
   • x = α - lγ/n
   • y = β - mγ/n
5. This intersection point (x, y, 0) must lie on the guiding curve f(x, y) = 0.
   f(α - lγ/n, β - mγ/n) = 0
6. This is the condition for the line to be a generator. Now, we eliminate the unknowns l, m, n using the original line equations (step 3).
   From (3): l/n = (x - α)/(z - γ) and m/n = (y - β)/(z - γ)
7. Substitute these into the condition from step 5:

   f( α - (x - α)γ/(z - γ), β - (y - β)γ/(z - γ) ) = 0

8. Simplifying the terms inside:
   • α - (x - α)γ/(z - γ) = (αz - αγ - xγ + αγ)/(z - γ) = (αz - xγ)/(z - γ)
   • β - (y - β)γ/(z - γ) = (βz - βγ - yγ + βγ)/(z - γ) = (βz - yγ)/(z - γ)

The final equation of the cone is:

f( (αz - γx)/(z - γ), (βz - γy)/(z - γ) ) = 0

Example:

Question: Find the equation of the cone with vertex (α, β, γ) and base ax² + by² = 1, z = 0.

Solution:
Here f(x, y) = ax² + by² - 1 = 0.
Substituting the derived terms into f(..., ...) = 0:
a[ (αz - γx)/(z - γ) ]² + b[ (βz - γy)/(z - γ) ]² - 1 = 0

a(αz - γx)² + b(βz - γy)² = (z - γ)²

This is the required equation.

Right Circular Cone

Definition

A **right circular cone** is a special type of cone where:

1. The guiding curve (base) is a circle.
2. The vertex lies on the line (the **axis**) that passes through the center of the circle and is perpendicular to the plane of the circle.

All generators make a constant angle θ with the axis. This angle θ is called the **semi-vertical angle**.

Equation of a Right Circular Cone

The equation is derived from the constant angle property.
Let the cone have:

• Vertex: V(α, β, γ)
• Axis direction cosines: (l, m, n) (Note: l² + m² + n² = 1)
• Semi-vertical angle: θ

Let P(x, y, z) be any point on the cone. The line segment VP is a generator.
The direction ratios of the generator VP are (x - α, y - β, z - γ).
The angle between the generator VP and the axis must be θ.

Using the angle formula cos(θ) = (a₁a₂ + b₁b₂ + c₁c₂) / (sqrt(a₁² + ...) * sqrt(a₂² + ...)):

cos(θ) = [ l(x - α) + m(y - β) + n(z - γ) ] / [ sqrt(l²+m²+n²) * sqrt((x-α)² + (y-β)² + (z-γ)²) ]

Since sqrt(l²+m²+n²) = 1, we can square both sides to get the final equation:

[ l(x - α) + m(y - β) + n(z - γ) ]² = cos²(θ) * [ (x - α)² + (y - β)² + (z - γ)² ]

The Cylinder

Definition

A **cylinder** is a 3D surface generated by a moving straight line (the **generator**) which remains parallel to a fixed line and intersects a given curve (the **guiding curve** or **base**).

Equation of a Cylinder

The method is similar to the cone, but the generator equations are different.

Method:

1. Let the generators be parallel to the fixed line x/l = y/m = z/n.
2. Let the guiding curve (base) be f(x, y) = 0 in the plane z = 0.
3. Let P(x₁, y₁, z₁) be *any* point on the cylinder.
4. The generator passing through P has equations:
   (x - x₁)/l = (y - y₁)/m = (z - z₁)/n
5. This generator must intersect the base. Find the point where it intersects z = 0.
   Set z = 0:
   (x - x₁)/l = (y - y₁)/m = (0 - z₁)/n
   Solve for x and y:
   • x = x₁ - lz₁/n
   • y = y₁ - mz₁/n
6. This intersection point (x, y, 0) must lie on the guiding curve f(x, y) = 0.
   f(x₁ - lz₁/n, y₁ - mz₁/n) = 0
7. This is the condition for (x₁, y₁, z₁) to be on the cylinder. To get the locus, replace (x₁, y₁, z₁) with (x, y, z).

The final equation of the cylinder is:

f( x - lz/n, y - mz/n ) = 0
or
f( (nx - lz)/n, (ny - mz)/n ) = 0

Example:

Question: Find the equation of the cylinder whose generators are parallel to x = y = z and whose base is the ellipse x² + 2y² = 1, z = 0.

Solution:
1. The fixed line is x/1 = y/1 = z/1. So, l=1, m=1, n=1.
2. The base is f(x, y) = x² + 2y² - 1 = 0.
3. The locus is f(x - lz/n, y - mz/n) = 0.
4. Substitute values: f(x - 1z/1, y - 1z/1) = 0 => f(x - z, y - z) = 0.
5. Apply this to f:

(x - z)² + 2(y - z)² = 1

This is the required equation.

Right Circular Cylinder

Definition

A **right circular cylinder** is a cylinder where the guiding curve is a circle and the generators are perpendicular to the plane of the circle. The fixed line to which all generators are parallel is called the **axis** of the cylinder.

Alternatively, a right circular cylinder is the locus of a point in 3D space that is at a constant distance (the **radius**) from a fixed line (the **axis**).

Equation of a Right Circular Cylinder

This locus definition gives the easiest way to find the equation.

Let the cylinder have:

• Axis: The line passing through A(α, β, γ) with direction cosines (l, m, n).
• Radius: r

Let P(x, y, z) be any point on the cylinder. The perpendicular distance from P to the axis must be equal to r.

We use a formula from 3D geometry: The perpendicular distance d from a point P to a line through A with d.c.s (l,m,n) is given by:

d² = (AP)² - (Projection of AP on the line)²

• (AP)² = (x - α)² + (y - β)² + (z - γ)²
• Projection of AP = l(x - α) + m(y - β) + n(z - γ)

Setting d² = r², we get the equation of the cylinder:

r² = [ (x - α)² + (y - β)² + (z - γ)² ] - [ l(x - α) + m(y - β) + n(z - γ) ]²

Unit 5: Exam Quick Tips