Degenerate Conic Sections

In analytic geometry, a degenerate conic section (or degenerate conic) is one that fails to exhibit the standard properties that characterize conic sections. This occurs when the determinant associated with the conic vanishes (δ = 0), causing the conic to reduce to simpler geometric figures such as a point or lines.

From a geometric perspective, a degenerate conic section can be visualized as the intersection of a plane with a double-napped right circular cone when the plane passes through the vertex of the cone.

Degenerate conic sections are classified into three main types:

• Point: when the plane intersects the cone only at the vertex without cutting through either nappe.
• Line (two coincident lines): when the plane is tangent to the cone along a generator and passes through the vertex. Because the two lines coincide, they appear as a single line.
• Pair of intersecting lines: when the plane passes through the vertex and cuts through both nappes of the cone.

Each conic section has its corresponding degenerate cases, as shown in the following table:

Conic Section	Degenerate Case	Degenerate Equation
Ellipse	A single real point	(x - h)² + (y - k)² = 0 The graph is the point (h, k)
Hyperbola	Two intersecting lines forming an X.	(x - h)² / a² - (y - k)² / b² = 0 (y - k)² / a² - (x - h)² / b² = 0 The graph consists of two lines intersecting at (h, k).
Parabola	Two parallel or coincident lines, real or imaginary.	(x - h)² = a (y - k)² = a When a > 0, the lines are real and parallel. When a < 0, the lines are imaginary. When a = 0, the lines are coincident.

Examples

Degenerate Ellipses

The following are equations of ellipses that degenerate into a single point.

• \(x^2+y^2=0\)
• \((y-3)^2+(x+1)^2=0\)
• \(x^2+(y+6)^2=0\)
• \(-(x+4)^2-y^2=0\)
• \((x-2)^2=-(y+4)^2\)

An equation of the form \((x-h)^2+(y-k)^2=c\), where c is negative, corresponds to an imaginary ellipse rather than a degenerate case. For instance, the equation \(x^2+y^2=-1\) represents an ellipse with no real points. The following are graphs of degenerate ellipses:

Degenerate Hyperbolas

The following are equations of hyperbolas that degenerate into two intersecting lines.

• \(x^2-y^2=0\)
• \((x+5)^2-(y+1)^2 = 0\)
• \(\dfrac{(x-1)^2}{4}-\dfrac{(y+3)^2}{9}=0\)
• \(\dfrac{(y-2)^2}{16}-\dfrac{(x+5)^2}{25}=0\)
• \(\dfrac{(x+3)^2}{1}=\dfrac{(y-4)^2}{2}\)

An equation of the form \(\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=0\) represents two intersecting lines whose slopes are b/a and -b/a. The following are graphs of degenerate hyperbolas:

Degenerate Parabolas

The following are equations of parabolas that degenerate into parallel or coincident lines. These can be real or imaginary.

• \(x^2=0\) (coincident lines)
• \(y^2=-6\) (imaginary lines)
• \((y-2)^2=3\)
• \((x+1)^2=0\)
• \(y^2-3y=-2\)
• \(x^2=5\)

In equations of the form (x-h)² = a or (y-k)² = a, the parallel lines are separated by a distance \(2 \sqrt{a}.\) The following are graphs of degenerate parabolas:

Note: In the general second-degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0, if A, B, and C are all zero, the formula reduces to Dx + Ey + F = 0, which represents a straight line in the plane. However, this type of linear equation does not represent a conic section, and therefore is not a degenerate case. For an equation of the form Ax² + Bxy + Cy² + Dx + Ey + F = 0 to represent a conic section (either non-degenerate or degenerate), A, B, and C cannot all be zero simultaneously—that is, at least one of these coefficients must be non-zero, which does not hold for Dx + Ey + F = 0.

How to Identify a Degenerate Conic Section

There are two primary methods for determining whether an equation represents a degenerate conic section: an algebraic approach and a matrix invariant approach.

Analyzing the Equation

The first method consists of analyzing the conic's equation directly. If the xy-term does not appear, you can complete the square and determine whether the resulting standard form corresponds to a degenerate case. On the other hand, if the xy-term is present, you can factor the expression or solve for one variable in terms of the other using the quadratic formula to reveal its geometric nature.

Example 1

Determine the nature of the equation \(2x^2+7xy+3y^2+x-7y-6=0.\)

Solution: To analyze the curve, treat the equation as a quadratic in x and apply the quadratic formula. Rewriting the terms gives:

$$2x^2+(7y+1)x+(3y^2-7y-6)=0$$

$$x=\dfrac{-(7y+1)\pm\sqrt{(7y+1)^2-4\cdot2\cdot(3y^2-7y-6)}}{4}$$

$$x=\dfrac{-(7y+1)\pm\sqrt{49y^2+14y+1-24y^2+56y+48}}{4}$$

$$x=\dfrac{-(7y+1)\pm\sqrt{25y^2+70y+49}}{4}$$

$$x=\dfrac{-(7y+1)\pm(5y+7)}{4}$$

Therefore,

$$x=\dfrac{-y+3}{2}\quad \text{and}\quad x=-3y-2$$

are the solutions. This means the given equation can be written in factored form as:

$$(2x+y-3)(x+3y+2)=0$$

If the coordinates of a point make either factor equal to zero, the entire product is zero, satisfying the original equation. Consequently, the graph consists of two intersecting lines whose equations are:

$$2x+y-3=0\quad \text{and}\quad x+3y+2=0$$

This is a degenerate hyperbola.

Example 2

Analyze the equation \(y^2-4xy+4x^2-2x+y-12=0.\)

Solution: Solving the equation for y in terms of x yields the two linear equations:

$$y-2x-3=0\quad \text{and}\quad y-2x+4=0$$

Consequently, the graph of the given equation consists of two parallel lines, representing a degenerate parabola.

Determinant Analysis

The second method for identifying a degenerate conic involves analyzing the equation using the invariant method. The general second-degree equation representing any conic section, whether regular or degenerate, has the form: Ax² + Bxy + Cy² + Dx + Ey + F = 0.

To determine whether the conic is non-degenerate or degenerate, compute the determinant of the matrix associated with the quadratic equation:

$$δ = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \\ \end{vmatrix}$$

Expanding this determinant yields:

$$δ = 8ACF-2AE^2-2B^2F+2BED-2CD^2$$

The value of this determinant indicates whether the conic is non-degenerate or degenerate:

• If δ ≠ 0: the conic is non-degenerate (regular), meaning the equation represents a standard smooth curve such as a circle, ellipse, parabola, or hyperbola.
• If δ = 0: the conic is degenerate, meaning the equation represents a point, a pair of intersecting lines, or a pair of parallel or coincident lines.

Once a conic is found to be degenerate (δ = 0), the next step is to classify the specific type of degeneration using the discriminant:

Δ = B² - 4AC

This value classifies the degenerate conic as follows:

• B² - 4AC > 0: the conic is a degenerate hyperbola, which manifests as a pair of real intersecting lines.
• B² - 4AC < 0: the conic is a degenerate ellipse, which appears as a single real point in the plane.
• B² - 4AC = 0: the conic is a degenerate parabola, which appears as two parallel or coincident lines (either real or imaginary).

In summary, the procedure to determine whether an equation of the form Ax² + Bxy + Cy² + Dx + Ey + F = 0 represents a degenerate conic is:

1. Compute the determinant \(δ = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \\ \end{vmatrix}.\) If the result is zero, the conic is degenerate; if it is non-zero, the conic is regular.
2. Compute the discriminant Δ = B² - 4AC. If positive, it corresponds to a degenerate hyperbola; if negative, to a degenerate ellipse; if zero, to a degenerate parabola.

Determinant	Discriminant	Conic Section
δ = 0	B2 - 4AC > 0	Degenerate hyperbola: pair of real intersecting lines.
δ = 0	B2 - 4AC < 0	Degenerate ellipse: a single real point.
δ = 0	B2 - 4AC = 0	Degenerate parabola: two parallel or coincident lines (real or imaginary).

Example 1

Analyze the equation \(y^2+2xy+x^2-2x-2y=0.\)

Solution: Begin by identifying the coefficients corresponding to the general conic form \(Ax^2+Bxy+Cy^2+Dx+Ey+F=0.\) Here, the coefficients are A=1, B=2, C=1, D=-2, E=-2, and F=0.

First, calculate the determinant of the matrix associated with the quadratic equation to determine whether the conic is regular or degenerate:

$$\delta=\begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix}=\begin{vmatrix} 2(1) & 2 & -2 \\ 2 & 2(1) & -2 \\ -2 & -2 & 2(0) \end{vmatrix}$$

Expanding the determinant:

$$\delta=8\cdot1\cdot0-2\cdot1\cdot(-2)^2-2\cdot2^2\cdot0+2\cdot2\cdot(-2)\cdot(-2)-2\cdot1\cdot(-2)^2=0$$

Since δ = 0, the conic is degenerate. Next, identify the type of degeneration by computing the discriminant:

$$\Delta=B^2-4AC=2^2-4(1)(1)=4-4=0$$

Because Δ = 0, the conic is classified as a degenerate parabola. This means the equation represents two parallel or coincident lines.

Example 2

Determine whether the equation \(2x^2-xy-y^2+2x+y=0\) represents a regular or degenerate conic, and determine its type.

Solution: First, identify the coefficients in the general form \(Ax^2+Bxy+Cy^2+Dx+Ey+F=0.\) Here, A=2, B=-1, C=-1, D=2, E=1, and F=0.

Calculate the determinant of the matrix associated with the quadratic equation:

$$\delta= \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix}=\begin{vmatrix} 2(2) & -1 & 2 \\ -1 & 2(-1) & 1 \\ 2 & 1 & 2(0) \end{vmatrix}$$

Expanding the determinant:

$$\delta=8\cdot(-2)\cdot0-2\cdot2\cdot1^2-2\cdot(-1)^2\cdot0+2\cdot(-1)\cdot1\cdot2-2\cdot(-1)\cdot2^2$$

Simplifying:

$$\delta=0-4-0-4+8=0$$

Since δ = 0, the conic is degenerate. To classify the type of degeneration, compute the discriminant:

$$\Delta=B^2-4AC=(-1)^2-4(2)(-1)=1+8=9$$

Because Δ = 9 > 0, the conic is a degenerate hyperbola, which manifests as a pair of real intersecting lines.

Example 3

Analyze the equation \(25x^2 + 36y^2 + 30xy = 0.\)

Solution: Begin by identifying the coefficients in the general form: A=25, B=30, C=36, D=0, E=0, and F=0.

Calculate the determinant of the matrix associated with the conic to verify whether it is regular or degenerate:

$$\delta=\begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix} =\begin{vmatrix} 2(25) & 30 & 0 \\ 30 & 2(36) & 0 \\ 0 & 0 & 2(0) \end{vmatrix}$$

Expanding this determinant:

$$\delta=(50)(72)(0)-2(25)(0)^2-2(30)^2(0)+2(30)(0)(0)-2(36)(0)^2=0$$

Simplifying gives δ = 0, indicating that the conic is degenerate. Next, evaluate the type of degeneration by calculating the discriminant:

$$\Delta=B^2-4AC=(30)^2-4(25)(36)=900-3600=-2700$$

Because Δ = -2700 < 0, the conic is a degenerate ellipse. This type of degeneration appears as a single point in the plane.