Unit 5: Cones and Cylinders
Rigorous investigation of quadric ruled surfaces: homogeneous cones with vertex at origin, conditions for general quadric cones, right circular cones and cylinders, and enveloping surfaces.
ยง5.1 5.1 Cones with Vertex at the Origin and General Conic Surfaces
1. Geometric Definition of a Cone
A cone is a surface generated by a moving straight line (called a generator) which passes through a fixed point (called the vertex) and continually intersects a fixed space curve (called the guiding curve).
2. Theorem: Cones with Vertex at the Origin
Theorem: Every homogeneous polynomial equation of degree two in $x, y, z$:
represents a cone whose vertex is at the origin $(0, 0, 0)$.
Proof:
Let $P(x_1, y_1, z_1)$ be any point lying on the surface, so that:
Consider the straight line connecting the origin $O(0, 0, 0)$ to $P(x_1, y_1, z_1)$. The coordinates of any arbitrary point $Q$ on this line are given by $(t x_1, t y_1, t z_1)$ for any scalar parameter $t \in \mathbb{R}$.
Substitute the coordinates of $Q$ into the left-hand side of the quadratic equation:
By equation (1), the bracketed expression is identically zero. Therefore:
This proves that every point on the line $OP$ lies entirely on the surface. Since this holds for every point $P$ on the surface, the surface is generated by a family of straight lines all passing through the origin. Hence, it is a cone with vertex $(0, 0, 0)$. $\blacksquare$
3. General Second-Degree Equation Representing a Cone
Consider the general quadratic equation:
If this represents a cone with vertex $V(\alpha, \beta, \gamma)$, shifting the origin to $(\alpha, \beta, \gamma)$ via $x = X + \alpha$, $y = Y + \beta$, $z = Z + \gamma$ must eliminate all linear and constant terms, yielding a homogeneous quadratic in $X, Y, Z$.
This requirement is equivalent to the simultaneous vanishing of all partial derivatives at the vertex:
For these four linear equations in $(\alpha, \beta, \gamma, 1)$ to have a consistent non-trivial solution, the $4 \times 4$ determinant of coefficients must vanish:
This is the necessary and sufficient condition for the general quadratic equation to represent a cone.
ยง5.2 5.2 The Right Circular Cone
1. Geometric Definition
A right circular cone is a surface generated by a line passing through a fixed vertex, maintaining a constant angle $\alpha$ (the semi-vertical angle) with a fixed line through the vertex (the axis).
2. Analytical Derivation of the Equation
Let:
- Vertex: $V(\alpha_v, \beta_v, \gamma_v)$
- Axis: Line passing through $V$ with direction cosines $(l, m, n)$ (where $l^2 + m^2 + n^2 = 1$)
- Semi-vertical angle: $\alpha$ ($0 < \alpha < \frac{\pi}{2}$)
Let $P(x, y, z)$ be any point on the cone. The generator line $VP$ has direction ratios:
The cosine of the angle between $\vec{VP}$ and the axis vector $\vec{A} = (l, m, n)$ is $\cos\alpha$:
Squaring both sides and setting $l^2 + m^2 + n^2 = 1$:
Canonical Case (Vertex at Origin, Axis along Z-Axis):
If $V = (0, 0, 0)$ and the axis is the $z$-axis $(l=0, m=0, n=1)$:
This is the familiar standard equation of a right circular cone.
ยง5.3 5.3 Cylinders and Enveloping Cones/Cylinders
1. The General Cylinder
A cylinder is a surface generated by a straight line (the generator) that moves parallel to a fixed given direction and continually intersects a fixed guiding curve.
Equation with Generators Parallel to Given Line:
Let the generators have direction ratios $(l, m, n)$, and let the guiding curve be a plane conic in $z = 0$:
Let $P(x, y, z)$ be any point on the cylinder. The generator through $P$ has equations:
This line pierces the plane $Z = 0$ when:
Since $(X, Y, 0)$ lies on the guiding curve $f(X, Y) = 0$, substitute $X$ and $Y$:
Expanding this gives the Cartesian equation of the cylinder.
2. Right Circular Cylinder
A cylinder whose guiding curve is a circle and whose generators are perpendicular to the plane of the circle is a right circular cylinder.
If the axis is the line $\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$ (with $l^2+m^2+n^2=1$) and radius is $r$, the perpendicular distance from any point $P(x, y, z)$ to the axis equals $r$:
3. Enveloping Cone and Enveloping Cylinder of a Sphere
Enveloping Cone:
The locus of tangent lines drawn from an external point $V(x_1, y_1, z_1)$ to a sphere $S: x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$ is called the enveloping cone.
By Joachimsthal's quadratic method, if a line through $V$ touches the sphere, the quadratic equation for the intersection points has equal roots (discriminant $\Delta = 0$). This yields the fundamental relation:
where:
- $S = x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d$
- $S_1 = x_1^2 + y_1^2 + z_1^2 + 2ux_1 + 2vy_1 + 2wz_1 + d$
- $T = x x_1 + y y_1 + z z_1 + u(x + x_1) + v(y + y_1) + w(z + z_1) + d$
Enveloping Cylinder:
Similarly, the locus of tangent lines to a sphere $S = 0$ parallel to a fixed direction $(l, m, n)$ is the enveloping cylinder of the sphere.
Step-by-Step Solved Examination Problems
Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.
Find the equation of the right circular cone whose vertex is at the point $V(1, 2, 3)$, whose axis has direction ratios $(2, -1, 2)$, and whose semi-vertical angle is $\alpha = 45^\circ$.
Step 1: Compute the direction cosines of the axis The direction ratios are $(2, -1, 2)$. The magnitude is:
Direction cosines: $(l, m, n) = \left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)$.
Step 2: Set up the angle condition Let $P(x, y, z)$ be any point on the cone. The vector from vertex $V(1, 2, 3)$ to $P$ is:
The cosine of the angle between $\vec{VP}$ and the axis is $\cos 45^\circ = \frac{1}{\sqrt{2}}$:
Multiplying both sides by 3:
Step 3: Square both sides
Multiply by 2:
This is the exact Cartesian equation of the right circular cone.
Find the equation of the cylinder whose generators are parallel to the line:
and whose guiding curve is the ellipse in the $xy$-plane given by:
Step 1: Set up the generator line through an arbitrary point Let $P(x, y, z)$ be an arbitrary point on the cylinder. The generators have direction ratios $(l, m, n) = (1, 2, 3)$.
The equations of the generator line passing through $P(x, y, z)$ are:
Step 2: Find the point of intersection with $Z = 0$ Setting $Z = 0$:
Solving for $X$ and $Y$:
Step 3: Impose the guiding curve condition The intersection point $(X, Y, 0)$ must satisfy the ellipse equation $X^2 + 2Y^2 = 1$:
Expanding:
Multiply through by 9:
Expanding each squared term:
Dividing the entire equation by 3:
This is the exact Cartesian equation of the cylinder.
Find the equation of the enveloping cone of the sphere:
with vertex at the external point $V(1, 1, 1)$. Verify that the vertex satisfies the tangency envelope condition.
Step 1: Write the sphere equation and compute $S_1$ and $T$ The sphere is $S(x, y, z) = x^2 + y^2 + z^2 - 2x + 4y - 1 = 0$.
For vertex $V(x_1, y_1, z_1) = (1, 1, 1)$:
Now evaluate the tangent form $T(x, y, z)$:
Substitute $x_1 = 1, y_1 = 1, z_1 = 1$:
Step 2: Apply the Joachimsthal Envelope Formula $S S_1 = T^2$
Expand both sides:
Step 3: Collect all terms on one side
Step 4: Verification of Vertex Check if $V(1, 1, 1)$ lies on this quadric:
Thus, the equation of the enveloping cone is: