Chapter 6 • Theory & Derivations

Unit 6: Central and Non-Central Conicoids

Comprehensive classification and properties of quadric surfaces: ellipsoids, hyperboloids of one and two sheets, elliptic and hyperbolic paraboloids, tangent planes, polar planes, and director spheres.

§6.1 6.1 Canonical Classification of Central Quadrics

A conicoid (or quadric surface) is the locus of an equation of the second degree in three variables $x, y, z$. When a center of symmetry exists, the surface is called a central conicoid.

By an appropriate choice of orthogonal coordinate axes aligned with the principal axes of symmetry, every central conicoid can be reduced to the canonical form:

$$A x^2 + B y^2 + C z^2 = 1$$

where $a, b, c > 0$ are the semi-axes.

1. The Ellipsoid

\mathbf{\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1}

Geometry: A closed, bounded convex surface contained entirely within the box $[-a, a] \times [-b, b] \times [-c, c]$.
Planar Sections: Every planar section $z = k$ with $|k| < c$ is an ellipse:

\frac{x^2}{a^2(1 - k^2/c^2)} + \frac{y^2}{b^2(1 - k^2/c^2)} = 1

Volume: $V = \frac{4}{3}\pi abc$.

2. Hyperboloid of One Sheet

\mathbf{\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1}

Geometry: A connected, unbounded surface shaped like a cooling tower or hourglass.
Planar Sections:
Horizontal sections $z = k$ are ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + \frac{k^2}{c^2}$, expanding as $|k|$ increases. The smallest ellipse occurs at the waist ($z = 0$), known as the principal elliptic section.
Vertical sections $x = k$ or $y = k$ are hyperbolas.
Doubly Ruled Surface: Through every point on the hyperboloid of one sheet, there pass two distinct straight lines lying entirely within the surface! The two ruling families are:

\begin{cases} \frac{x}{a} - \frac{z}{c} = \lambda\left(1 - \frac{y}{b}\right) \\ \frac{x}{a} + \frac{z}{c} = \frac{1}{\lambda}\left(1 + \frac{y}{b}\right) \end{cases} \quad \text{and} \quad \begin{cases} \frac{x}{a} + \frac{z}{c} = \mu\left(1 - \frac{y}{b}\right) \\ \frac{x}{a} - \frac{z}{c} = \frac{1}{\mu}\left(1 + \frac{y}{b}\right) \end{cases}

3. Hyperboloid of Two Sheets

\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1} \quad \iff \quad \frac{x^2}{a^2} - 1 = \frac{y^2}{b^2} + \frac{z^2}{c^2}

Geometry: A disconnected surface consisting of two separate convex sheets separated by a gap of length $2a$ along the $x$-axis.
Regions: No real points exist for $|x| < a$. For $|x| \ge a$, planar sections perpendicular to the $x$-axis are ellipses.

§6.2 6.2 Non-Central Conicoids (Paraboloids)

When one of the quadratic terms vanishes and a linear term appears, the surface possesses no center of symmetry. These are the paraboloids.

1. The Elliptic Paraboloid

\mathbf{\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{2z}{c}} \quad (c > 0)

Geometry: A cup- or bowl-shaped surface extending infinitely in the positive $z$-direction ($z \ge 0$).
Planar Sections:
Horizontal slices $z = k > 0$ yield ellipses: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{2k}{c}$.
Vertical slices parallel to the $xz$-plane ($y = k_y$) yield parabolas opening upwards: $z = \frac{c}{2a^2}x^2 + \text{const}$.
Vertical slices parallel to the $yz$-plane ($x = k_x$) also yield parabolas opening upwards.
Vertex: The origin $(0, 0, 0)$ is the unique vertex and absolute minimum.

2. The Hyperbolic Paraboloid (Saddle Surface)

\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2} = \frac{2z}{c}} \quad (c > 0)

Geometry: A saddle-shaped surface featuring a mountain pass or minimax point at the origin $(0, 0, 0)$.
Planar Sections:
Horizontal slices $z = k > 0$ yield hyperbolas opening along the $x$-axis.
Horizontal slices $z = k < 0$ yield hyperbolas opening along the $y$-axis.
The slice $z = 0$ yields a pair of intersecting straight lines: $\frac{x}{a} = \pm \frac{y}{b}$.
Vertical slices $y = k$ are parabolas opening upward: $z = \frac{c}{2a^2}x^2 - \frac{c k^2}{2b^2}$.
Vertical slices $x = k$ are parabolas opening downward: $z = -\frac{c}{2b^2}y^2 + \frac{c k^2}{2a^2}$.
Doubly Ruled Surface: Like the hyperboloid of one sheet, the hyperbolic paraboloid is a doubly ruled surface. Through every point on the saddle, there pass two straight lines lying entirely on the surface.

§6.3 6.3 Tangent Planes, Polar Planes, and Diametral Systems

1. Tangent Plane to a Central Conicoid

Let the central conicoid be:

$$A x^2 + B y^2 + C z^2 = 1$$

Let $P(x_1, y_1, z_1)$ be a point on the surface. The gradient vector at $P$ is:

\nabla F = (2A x_1, 2B y_1, 2C z_1)

The tangent plane at $P$ is perpendicular to $\nabla F$:

$$A x_1 (x - x_1) + B y_1 (y - y_1) + C z_1 (z - z_1) = 0$$

$$A x x_1 + B y y_1 + C z z_1 = A x_1^2 + B y_1^2 + C z_1^2$$

Since $(x_1, y_1, z_1)$ lies on the conicoid, the right-hand side equals 1:

\mathbf{A x x_1 + B y y_1 + C z z_1 = 1}

2. Condition of Tangency of a Plane

Let a plane be given by:

$$l x + m y + n z = p$$

If this plane is tangent to $A x^2 + B y^2 + C z^2 = 1$ at $(x_1, y_1, z_1)$, its coefficients must be proportional to the tangent plane equation:

\frac{A x_1}{l} = \frac{B y_1}{m} = \frac{C z_1}{n} = \frac{1}{p}

Solving for the contact point:

x_1 = \frac{l}{A p}, \quad y_1 = \frac{m}{B p}, \quad z_1 = \frac{n}{C p}

Substituting these coordinates into the conicoid equation $A x_1^2 + B y_1^2 + C z_1^2 = 1$:

A \left(\frac{l}{A p}\right)^2 + B \left(\frac{m}{B p}\right)^2 + C \left(\frac{n}{C p}\right)^2 = 1

\frac{l^2}{A p^2} + \frac{m^2}{B p^2} + \frac{n^2}{C p^2} = 1

\mathbf{\frac{l^2}{A} + \frac{m^2}{B} + \frac{n^2}{C} = p^2}

For an ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ ($A = 1/a^2, B = 1/b^2, C = 1/c^2$), the tangency condition is:

\mathbf{a^2 l^2 + b^2 m^2 + c^2 n^2 = p^2}

3. The Director Sphere

Theorem: The locus of the point of intersection of three mutually perpendicular tangent planes to a central conicoid $A x^2 + B y^2 + C z^2 = 1$ is a concentric sphere called the director sphere:

\mathbf{x^2 + y^2 + z^2 = \frac{1}{A} + \frac{1}{B} + \frac{1}{C}}

For the canonical ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$:

\mathbf{x^2 + y^2 + z^2 = a^2 + b^2 + c^2}

TIERED UNIVERSITY HONORS PROBLEMS

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.

Solved Problem Example 6.1: Tangent Plane and Normal Line to an Ellipsoid

Find the Cartesian equations of the tangent plane and the normal line to the ellipsoid:

\frac{x^2}{12} + \frac{y^2}{6} + \frac{z^2}{3} = 1

at the point $P(2, 1, 1)$.

Step 1: Verify the point lies on the ellipsoid

\frac{2^2}{12} + \frac{1^2}{6} + \frac{1^2}{3} = \frac{4}{12} + \frac{1}{6} + \frac{1}{3} = \frac{1}{3} + \frac{1}{6} + \frac{1}{3} = \frac{5}{6} \neq 1

Notice that $\frac{4}{12} + \frac{1}{6} + \frac{1}{3} = \frac{2}{6} + \frac{1}{6} + \frac{2}{6} = \frac{5}{6}$. Let us use the point $P(2, 1, \sqrt{3/2})$ or verify directly for $P(2, 1, 1)$ with the equation $\frac{x^2}{12} + \frac{y^2}{6} + \frac{z^2}{2} = 1$:

\frac{4}{12} + \frac{1}{6} + \frac{1}{2} = \frac{1}{3} + \frac{1}{6} + \frac{1}{2} = \frac{2 + 1 + 3}{6} = 1 \quad \checkmark

Let the ellipsoid be:

\frac{x^2}{12} + \frac{y^2}{6} + \frac{z^2}{2} = 1

at $P(2, 1, 1)$.

Step 2: Tangent Plane Equation The formula for the tangent plane at $(x_1, y_1, z_1)$ is:

\frac{x x_1}{a^2} + \frac{y y_1}{b^2} + \frac{z z_1}{c^2} = 1

Substitute $x_1 = 2, y_1 = 1, z_1 = 1$:

\frac{x(2)}{12} + \frac{y(1)}{6} + \frac{z(1)}{2} = 1

\frac{x}{6} + \frac{y}{6} + \frac{z}{2} = 1

Multiply through by 6:

$$x + y + 3z = 6$$

Step 3: Normal Line Equation The normal line passes through $(2, 1, 1)$ and has direction ratios equal to the normal vector of the tangent plane, which are $(1, 1, 3)$:

\frac{x - 2}{1} = \frac{y - 1}{1} = \frac{z - 1}{3}

Solved Problem Example 6.2: Tangent Planes Parallel to a Given Plane

Find the equations of the two tangent planes to the ellipsoid:

$$2x^2 + 6y^2 + 3z^2 = 6$$

that are parallel to the plane:

$$2x - 3y + 3z = 0$$

Also find the perpendicular distance between these two parallel tangent planes.

Step 1: Write ellipsoid in canonical form Divide $2x^2 + 6y^2 + 3z^2 = 6$ by 6:

\frac{x^2}{3} + \frac{y^2}{1} + \frac{z^2}{2} = 1

Here $a^2 = 3, b^2 = 1, c^2 = 2$.

Step 2: Set up parallel planes Any plane parallel to $2x - 3y + 3z = 0$ has the form:

$$2x - 3y + 3z = p$$

Here direction numbers are $l = 2, m = -3, n = 3$.

Step 3: Apply the condition of tangency For the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ and plane $l x + m y + n z = p$:

$$p^2 = a^2 l^2 + b^2 m^2 + c^2 n^2$$

Substitute values:

$$p^2 = 3(2^2) + 1(-3)^2 + 2(3^2)$$

$$p^2 = 3(4) + 1(9) + 2(9) = 12 + 9 + 18 = 39$$

Therefore:

p = \pm \sqrt{39}

The two tangent planes are:

2x - 3y + 3z = \sqrt{39} \quad \text{and} \quad 2x - 3y + 3z = -\sqrt{39}

Step 4: Distance between the parallel planes The distance between $2x - 3y + 3z = \sqrt{39}$ and $2x - 3y + 3z = -\sqrt{39}$ is:

D = \frac{|p_1 - p_2|}{\sqrt{l^2 + m^2 + n^2}} = \frac{2\sqrt{39}}{\sqrt{2^2 + (-3)^2 + 3^2}} = \frac{2\sqrt{39}}{\sqrt{4 + 9 + 9}} = \frac{2\sqrt{39}}{\sqrt{22}}

D = \sqrt{\frac{78}{11}} \approx 2.66 \text{ units}

Solved Problem Example 6.3: Director Sphere of a Central Conicoid

Prove analytically that the locus of the point of intersection of three mutually perpendicular tangent planes to the central conicoid:

$$a x^2 + b y^2 + c z^2 = 1$$

is the concentric sphere:

x^2 + y^2 + z^2 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}

(known as the Director Sphere).

Step 1: Set up three mutually orthogonal planes Let $(x, y, z)$ be the point of intersection of three mutually perpendicular tangent planes.

Let the unit normal vectors of the three planes be:

\hat{n}_1 = (l_1, m_1, n_1), \quad \hat{n}_2 = (l_2, m_2, n_2), \quad \hat{n}_3 = (l_3, m_3, n_3)

Since the normals are mutually orthogonal unit vectors, the matrix $M$ formed by these vectors is orthogonal ($M M^T = I$). Consequently, the column-orthogonality relations hold:

\begin{cases} l_1^2 + l_2^2 + l_3^2 = 1 \ m_1^2 + m_2^2 + m_3^2 = 1 \ n_1^2 + n_2^2 + n_3^2 = 1 \ l_1 m_1 + l_2 m_2 + l_3 m_3 = 0 \ m_1 n_1 + m_2 n_2 + m_3 n_3 = 0 \ n_1 l_1 + n_2 l_2 + n_3 l_3 = 0 \end{cases}

Step 2: Express each tangent plane equation A plane with unit normal $(l_i, m_i, n_i)$ tangent to $a x^2 + b y^2 + c z^2 = 1$ is:

$$l_i x + m_i y + n_i z = p_i$$

where the tangency condition requires:

p_i^2 = \frac{l_i^2}{a} + \frac{m_i^2}{b} + \frac{n_i^2}{c}

Since $(x, y, z)$ is the intersection point, it satisfies all three equations:

$$p_1 = l_1 x + m_1 y + n_1 z$$

$$p_2 = l_2 x + m_2 y + n_2 z$$

$$p_3 = l_3 x + m_3 y + n_3 z$$

Step 3: Sum the squares of the equations

$$p_1^2 + p_2^2 + p_3^2 = (l_1 x + m_1 y + n_1 z)^2 + (l_2 x + m_2 y + n_2 z)^2 + (l_3 x + m_3 y + n_3 z)^2$$

Expanding the RHS and grouping coefficients:

\text{RHS} = x^2 (l_1^2 + l_2^2 + l_3^2) + y^2 (m_1^2 + m_2^2 + m_3^2) + z^2 (n_1^2 + n_2^2 + n_3^2)

$$+ 2xy (l_1 m_1 + l_2 m_2 + l_3 m_3) + 2yz (m_1 n_1 + m_2 n_2 + m_3 n_3) + 2zx (n_1 l_1 + n_2 l_2 + n_3 l_3)$$

Using the orthogonality relations from Step 1:

\text{RHS} = x^2 (1) + y^2 (1) + z^2 (1) + 0 + 0 + 0 = x^2 + y^2 + z^2

Step 4: Sum the LHS using the tangency condition

\text{LHS} = p_1^2 + p_2^2 + p_3^2 = \sum_{i=1}^3 \left( \frac{l_i^2}{a} + \frac{m_i^2}{b} + \frac{n_i^2}{c} \right)

= \frac{1}{a} \sum_{i=1}^3 l_i^2 + \frac{1}{b} \sum_{i=1}^3 m_i^2 + \frac{1}{c} \sum_{i=1}^3 n_i^2

= \frac{1}{a}(1) + \frac{1}{b}(1) + \frac{1}{c}(1) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}

Step 5: Equate LHS and RHS

x^2 + y^2 + z^2 = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}

This is a sphere centered at the origin $(0, 0, 0)$ with radius $R = \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$. $\blacksquare$