AP Precalculus • Functions Involving Parameters, Vectors, and Matrices

Conic Sections

The elegant family of curves formed by slicing a cone—parabolas, ellipses, hyperbolas—unified by parametric and matrix representations.

A function curve with a triangle and circular angle measure.
Inside this lesson 9 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

The study of conic sections ranks among the oldest investigations in mathematics, originating in ancient Greece and persisting as one of the most consequential topics in analytic geometry. These curves—circles, ellipses, parabolas, and hyperbolas—arise naturally when a plane intersects a double-napped cone at various angles. Greek geometers initially studied conics for their intrinsic beauty, but the curves later proved indispensable for describing planetary orbits, projectile trajectories, and the behavior of light and sound. In the modern AP Precalculus curriculum, conic sections serve as a bridge between algebraic equations, parametric representations, and matrix transformations, making them a natural testing ground for the unit on parameters, vectors, and matrices.

~350 BCE • Menaechmus & the Double Cone

Menaechmus, a student of Plato, reportedly discovered conic sections while attempting to solve the Delian problem of doubling the cube, recognizing that parabolas and hyperbolas could yield the necessary geometric means.

~200 BCE • Apollonius of Perga

Apollonius wrote his eight-volume treatise Conics, introducing the names ellipse, parabola, and hyperbola and systematizing their properties using the double-napped cone framework still used today.

1609 • Kepler's Elliptical Orbits

Johannes Kepler published his first law of planetary motion, demonstrating that planets travel in elliptical orbits with the Sun at one focus—validating the physical significance of conics.

1637 • Descartes & Analytic Geometry

René Descartes unified algebra and geometry by representing curves—including all conics—as polynomial equations in a coordinate plane, enabling the algebraic classification used in modern courses.

The central question that conic sections answer is deceptively simple: how can a single family of second-degree equations describe such diverse shapes, and how do parametric equations and matrix operations let us manipulate, rotate, and translate these curves with precision? Mastering conics equips you with the algebraic and geometric fluency needed for calculus, physics, and engineering.

SECTION 2

Core Principles & Definitions

Every conic section can be understood as the set of points satisfying a specific geometric relationship involving distances to fixed elements—a focus (a fixed point) and a directrix (a fixed line). The ratio of the distance from any point on the curve to the focus versus its distance to the directrix is the eccentricity (e), and this single parameter determines the conic's type. Equivalently, conics are defined as the intersection of a plane with a right circular double cone—the angle of the cutting plane relative to the cone's axis determines which curve results.

1

Circle (e = 0)

All points equidistant from a single center. A circle is a special ellipse where both foci coincide, giving eccentricity exactly zero.
2

Ellipse (0 < e < 1)

The locus of points whose sum of distances to two foci is constant. Eccentricity between 0 and 1 produces a closed, oval curve.
3

Parabola (e = 1)

Each point is equidistant from the focus and the directrix. The curve is open and unbounded with exactly one branch.
4

Hyperbola (e > 1)

The locus of points whose absolute difference of distances to two foci is constant. Two separate branches open in opposite directions.

Sapphire double cone cut by the oblique plane z=0.5x+2, producing an ellipse. A view normal to the plane shows the true section shape.

Four cone-plane intersections: circle for z=2, ellipse for z=0.5x+2, parabola for z=x+2 and hyperbola for x=1.Four cone-plane intersections: circle for z=2, ellipse for z=0.5x+2, parabola for z=x+2 and hyperbola for x=1.

Horizontal plane z=2 cuts one cone nappe in a circle, with a plane-normal circular view.Horizontal plane z=2 cuts one cone nappe in a circle, with a plane-normal circular view.

Oblique plane z=0.5x+2 cuts one cone nappe in a closed ellipse, with a plane-normal view.Oblique plane z=0.5x+2 cuts one cone nappe in a closed ellipse, with a plane-normal view.

Plane z=x+2, parallel to a cone generator, gives an open parabola. The finite render boundary does not close the curve.Plane z=x+2, parallel to a cone generator, gives an open parabola. The finite render boundary does not close the curve.

Vertical plane x=1 cuts both cone nappes in two hyperbola branches, with a plane-normal view.Vertical plane x=1 cuts both cone nappes in two hyperbola branches, with a plane-normal view.

SECTION 3

Visualizing the Conic Family

The Four Conic Sections on a Coordinate PlaneCircle (e = 0)Ellipse (0 < e < 1)Parabola (e = 1)Hyperbola (e > 1)
Each conic is shown centered on its own coordinate axes. The circle (top-left) has a single center. The ellipse (top-right) features two foci. The parabola (bottom-left) has a single focus and a dashed directrix. The hyperbola (bottom-right) opens in two branches with two foci.

Notice how the geometric elements—center, foci, directrix, and axes of symmetry—shift as eccentricity changes. In the circle, the single center is the lone defining point. The ellipse introduces two foci that sit along the major axis; the closer they are together, the more circular the ellipse appears. The parabola's single focus and directrix work in tandem: every point on the curve is equidistant from both. Finally, the hyperbola's two branches curve away from each other, each approaching but never touching a pair of asymptotes. Being comfortable reading these visual features is essential before translating them into algebraic and parametric forms.

SECTION 4

Mathematical Framework

Standard Forms of Conics (Center at Origin)

GENERAL SECOND-DEGREE EQUATION

Ax2+Bxy+Cy2+Dx+Ey+F=0Ax² + Bxy + Cy² + Dx + Ey + F = 0

Every conic can be expressed in this form. The discriminant B² − 4AC classifies the conic: < 0 → ellipse/circle, = 0 → parabola, > 0 → hyperbola.

ELLIPSE (a > b)

x2/a2+y2/b2=1x²/a² + y²/b² = 1

Semi-major axis a along the x-axis, semi-minor axis b along the y-axis. Foci at (±c, 0) where c² = a² − b². When a = b, this becomes a circle of radius a.

HYPERBOLA (HORIZONTAL TRANSVERSE AXIS)

x2/a2−y2/b2=1x²/a² − y²/b² = 1

Opens left-right. Vertices at (±a, 0), foci at (±c, 0) where c² = a² + b². Asymptotes: y = ±(b/a)x.

PARABOLA (VERTICAL AXIS)

x2=4pyx² = 4py

Focus at (0, p), directrix at y = −p. If p > 0 the parabola opens upward; if p < 0 it opens downward. The analogous horizontal form is y² = 4px.

Parametric Representations

In the AP Precalculus context of parametric functions, conics can be traced by expressing x and y each as functions of a parameter t. For an ellipse with semi-axes a and b, the standard parametric form is x(t) = a cos t, y(t) = b sin t for t ∈ [0, 2π). For a hyperbola, we use x(t) = a sec t, y(t) = b tan t. These parametric forms are essential for eliminating the parameter to recover the rectangular equation, a core AP Precalculus skill. Rotation of conics by an angle θ can be handled by applying a rotation matrix to the parametric vector ⟨x(t), y(t)⟩, connecting this topic directly to the matrices strand of the course.

SECTION 5

Detailed Classification & the Discriminant

Given the general second-degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0, the value of the conic discriminant Δ = B² − 4AC provides a quick classification without completing the square or graphing. This quantity is invariant under rotation of axes, meaning it yields the same conic type regardless of orientation. The table below summarizes the classification alongside key geometric elements for each type.

ConicDiscriminant ΔEccentricityKey Relationship
Circle< 0 (and A = C, B = 0)e = 0All points equidistant from center
Ellipse< 0 (A ≠ C or B ≠ 0)0 < e < 1Sum of distances to foci = 2a
Parabola= 0e = 1Distance to focus = distance to directrix
Hyperbola> 0e > 1|Difference of distances to foci| = 2a

Classification of conics by discriminant and eccentricity

Eccentricity Spectrum: From Circle to Hyperbolae = 0Circle0 < e < 1Ellipsee = 1Parabolae > 1HyperbolaDiscriminant Quick-CheckΔ = B² − 4AC: negative → ellipse/circle · zero → parabola · positive → hyperbola(Degenerate cases excluded: e.g. point, line, or pair of intersecting lines)
The eccentricity spectrum at top shows how increasing e transitions from a circle through an ellipse, to a parabola, and finally a hyperbola. Below, representative curves are sketched for each conic type, and the discriminant rule is summarized in the shaded box.

SECTION 6

Worked Example: Identifying & Parametrizing an Ellipse

Consider the equation 9x² + 25y² − 54x + 100y − 44 = 0. We will classify this conic, rewrite it in standard form, identify its key features, and express it parametrically.

Worked example

  1. 1

    Step 1 — Apply the Discriminant Test

    Here A = 9, B = 0, C = 25. Compute Δ = B² − 4AC = 0 − 4(9)(25) = −900. Since Δ < 0 and A ≠ C, the conic is an ellipse.
    Δ = −900 → Ellipse
  2. 2

    Step 2 — Group and Complete the Square

    Group x-terms and y-terms: 9(x² − 6x) + 25(y² + 4y) = 44. Complete each square: 9(x² − 6x + 9) + 25(y² + 4y + 4) = 44 + 81 + 100 = 225. This yields 9(x − 3)² + 25(y + 2)² = 225.
    9(x − 3)² + 25(y + 2)² = 225
  3. 3

    Step 3 — Write in Standard Form

    Divide both sides by 225: (x − 3)²/25 + (y + 2)²/9 = 1. We read off a² = 25 so a = 5, b² = 9 so b = 3. The center is (h, k) = (3, −2). Since a > b and a is under the x-term, the major axis is horizontal.
    (x − 3)²/25 + (y + 2)²/9 = 1, center (3, −2)
  4. 4

    Step 4 — Find Foci and Eccentricity

    Compute c² = a² − b² = 25 − 9 = 16, so c = 4. Foci lie along the major axis at (3 ± 4, −2), giving (7, −2) and (−1, −2). Eccentricity e = c/a = 4/5 = 0.8.
    Foci: (7, −2) and (−1, −2); e = 0.8
  5. 5

    Step 5 — Write Parametric Equations

    Using x = h + a cos t and y = k + b sin t, we obtain x(t) = 3 + 5 cos t and y(t) = −2 + 3 sin t for t ∈ [0, 2π). We can verify by eliminating the parameter: cos t = (x − 3)/5 and sin t = (y + 2)/3, and since cos²t + sin²t = 1, we recover the standard form.
    x(t) = 3 + 5 cos t, y(t) = −2 + 3 sin t, t ∈ [0, 2π)

SECTION 7

Comparing Representations: Rectangular, Parametric, and Polar

Conics can be represented in rectangular (Cartesian) form, parametric form, and polar form. Each representation has strengths depending on the task—whether you need to graph, compute arc length, rotate the curve, or analyze orbital mechanics. Understanding when to convert between forms is a central AP Precalculus skill.

RepresentationStrengthsLimitations
RectangularDirect algebraic manipulation; discriminant classification; completing the square identifies center and axes immediately.Cannot represent direction of traversal; rotated conics produce messy Bxy cross-terms; not easily parameterized for calculus.
ParametricShows direction and speed of traversal; easy to apply rotation matrices; eliminates multi-valued y issues for vertical lines.Requires eliminating the parameter to recover the Cartesian equation; choice of parameter is not unique.
PolarElegantly unifies all conics via r = ed/(1 ± e cos θ); natural for orbital mechanics; eccentricity appears explicitly.Requires focus at the origin; not convenient for conics centered elsewhere without translation.

Comparison of conic representations

SECTION 8

Connection to Calculus, Matrices, and Beyond

Conic sections serve as a gateway to several advanced topics. In AP Calculus BC, parametric derivatives dy/dx = (dy/dt)/(dx/dt) allow you to find tangent lines and arc lengths of conics without ever solving for y explicitly. In linear algebra, the general second-degree equation corresponds to a quadratic form vTMv = 0, where M is a symmetric 2 × 2 matrix whose eigenvalues determine the conic type. Diagonalizing M is equivalent to rotating axes to eliminate the Bxy cross-term—an elegant connection to the matrices strand of this AP unit.

AP Precalculus TopicAdvanced Extension
Parametric equations of conicsParametric derivatives, arc length, and surface area of revolution in Calculus BC
Discriminant B² − 4ACEigenvalues of the associated matrix; quadratic form classification in linear algebra
Rotation of conic axesOrthogonal diagonalization; principal axis theorem
Polar form r = ed/(1 ± e cos θ)Kepler's laws; gravitational orbits; astrophysics trajectory calculations

From AP Precalculus to advanced topics

For now, focus on building fluency with the standard forms, parametric representations, and the discriminant test. These skills form the foundation upon which all advanced treatments—whether in multivariable calculus, differential equations, or physics—are built.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which conic section is described by the equation 4x² + 4y² − 16x + 24y − 36 = 0?
  • A. Parabola
  • B. Ellipse (not a circle)
  • C. Circle
  • D. Hyperbola
  • E. Degenerate conic (a single point)
Show answer (C)
Here A = 4, B = 0, C = 4. Since B = 0 and A = C, the discriminant Δ = 0 − 4(4)(4) = −64 < 0 with A = C, confirming this is a circle. Dividing through by 4 and completing the square: (x − 2)² + (y + 3)² = 22. Choice A (parabola) would require Δ = 0. Choice B (ellipse, not a circle) would require A ≠ C. Choice D (hyperbola) would require Δ > 0. Choice E (degenerate) would occur if the right side equaled zero or was negative after completing the square, but here the radius squared is 22 > 0.

Question 2: BASIC CALCULATION

An ellipse has the equation (x − 1)²/16 + (y + 3)²/9 = 1. What is the distance between its two foci?
  • A. √7
  • B. 2√7
  • C. 5
  • D. 7
  • E. 10
Show answer (B)
From the standard form, a² = 16 and b² = 9, so c² = a² − b² = 16 − 9 = 7, giving c = √7. The distance between the two foci is 2c = 2√7. Choice A gives only the distance from center to one focus (c = √7), not the total interfocal distance. Choice C (5) confuses c with the sum a + b = 4 + 3. Choice D (7) is c² rather than 2c. Choice E (10) equals 2a, the length of the major axis, not the distance between foci.

Question 3: INTERMEDIATE

A conic is given by 2x² − 3xy + y² + 4x − 2y + 1 = 0. Using the discriminant B² − 4AC, classify this conic and determine its eccentricity category.
  • A. Ellipse with e < 1
  • B. Parabola with e = 1
  • C. Hyperbola with e > 1
  • D. Circle with e = 0
  • E. Cannot be determined without eliminating the cross-term
Show answer (C)
Identify A = 2, B = −3, C = 1. Then Δ = (−3)² − 4(2)(1) = 9 − 8 = 1 > 0. A positive discriminant classifies this as a hyperbola with eccentricity e > 1. Choice A (ellipse) requires Δ < 0. Choice B (parabola) requires Δ = 0. Choice D (circle) requires A = C and B = 0. Choice E is incorrect because the discriminant is invariant under rotation—it classifies the conic without needing to eliminate the xy-term.

Question 4: APPLIED

A satellite orbits Earth in an elliptical path described parametrically by x(t) = 8000 cos t and y(t) = 6000 sin t (in km), where the center of the ellipse is at the origin. Determine the location of the foci (which represent possible positions for Earth's center) and write the rectangular equation of the orbit.
Show answer
From the parametric form, cos t = x/8000 and sin t = y/6000. Using the identity cos²t + sin²t = 1 gives x²/8000² + y²/6000² = 1, i.e., x²/64,000,000 + y²/36,000,000 = 1. Here a = 8000 and b = 6000, so c² = 64,000,000 − 36,000,000 = 28,000,000, giving c = 2000√7 ≈ 5291.5 km. The foci are at (±2000√7, 0). Earth's center would be located at one of these foci, consistent with Kepler's first law.

Question 5: CRITICAL THINKING

Consider the general conic Ax² + Bxy + Cy² = 1. (a) Show that the 2 × 2 matrix M = [[A, B/2],[B/2, C]] encodes the conic via the quadratic form vTv^T M v = 1, where v = [x, y]^T. (b) Explain why the eigenvalues of M being both positive corresponds to the conic being an ellipse. (c) Determine how the sign of det(M) relates to the discriminant B² − 4AC, and use this to justify the classification rule.
Show answer
(a) Writing v = [x, y]^T and computing vTv^T M v: [x y][[A, B/2],[B/2, C]][x, y]^T = Ax² + (B/2)xy + (B/2)yx + Cy² = Ax² + Bxy + Cy². Setting this equal to 1 recovers the original equation. (b) If both eigenvalues λ₁, λ₂ are positive, M is positive definite. The equation vTv^T M v = 1 defines a level curve of a strictly convex function, which must be a bounded, closed curve—an ellipse. If λ₁ = λ₂ the curve has equal 'radii' in the eigenvector directions and is a circle. (c) det(M) = AC − B²/4, so B² − 4AC = −4·det(M). When det(M) > 0 (both eigenvalues same sign, positive for a proper conic), Δ < 0 → ellipse. When det(M) = 0 (one eigenvalue is zero), Δ = 0 → parabola. When det(M) < 0 (eigenvalues of opposite sign), Δ > 0 → hyperbola. This provides the algebraic proof that the sign of the discriminant and the eigenvalue structure of the associated matrix yield identical classifications.

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