MGU Mathematics Series — Trigonometry and Precalculus
Part IV: Analytic Geometry, Parametric Forms, and Vectors
Circles · Parabolas · Ellipses · Hyperbolas · Applications
Chapter 11 of 14
MGU Mathematics Series  ·  Trigonometry and Precalculus  ·  Library Textbook Edition

Chapter 11:
Conic Sections and
Analytic Geometry

Circles, parabolas, ellipses, hyperbolas, completing the square, eccentricity, and applications in astronomy, optics, engineering, and design — the algebra of curves

■ Math Level: Trigonometry and Precalculus ■

■ 14 Sections  •  Part IV of 5 ■

Follows Chapter 10: Sequences, Series, and Mathematical Induction  ·  Prepares students for Chapter 12: Parametric Equations, Polar Coordinates, and Vectors

Chapter 11 — Front Matter

Purpose and Overview

Chapter 11 deepens the study of conic sections begun in Algebra II and connects it to the broader discipline of analytic geometry — the use of algebra to describe, analyze, and interpret geometric curves and shapes. Students will study the four classical conic sections: circles, parabolas, ellipses, and hyperbolas. They will learn to identify each curve from its equation, write equations from geometric descriptions, complete the square to convert equations into standard form, graph each conic with attention to key features, and interpret conics in applications drawn from astronomy, optics, engineering, architecture, and design.

This chapter also serves as a bridge into Chapter 12, where parametric equations and polar coordinates will offer additional languages for describing curves. Everything here is coordinate geometry: the meeting of algebra and geometry on the Cartesian plane.

Opening Question

A satellite dish focuses incoming signals at a single point called the focal point. An elliptical orbit carries a planet closer to and farther from the sun. A flashlight projects a cone of light that, when intercepted by a flat surface, creates a curved shape. What do all of these have in common? Each one involves a conic section — a curve formed by the intersection of a plane with a double cone. These four curves appear everywhere in the natural world, in engineering design, in optics, in physics, and in astronomy. Understanding their equations, their graphs, and their geometric properties is essential for any student preparing for calculus, physics, or applied mathematics.

14Sections
9Objectives
25+Key Terms
12+Worked Examples

Learning Objectives

  1. 1Explain what a conic section is and how it arises from the intersection of a plane and a double cone.
  2. 2Write and interpret the standard form equation of a circle, parabola, ellipse, and hyperbola.
  3. 3Identify key features of each conic: center, radius, vertex, focus, directrix, axes, asymptotes, and eccentricity.
  4. 4Complete the square to convert a general second-degree equation into standard conic form.
  5. 5Determine which conic a given equation represents from its algebraic structure.
  6. 6Graph all four conics with labeled key features.
  7. 7Apply conic equations to problems in astronomy, optics, engineering, and design.
  8. 8Explain the geometric meaning of eccentricity.
  9. 9Connect the focus-directrix definition of a parabola to its algebraic equation.
Chapter 11 — Contents

Sections at a Glance

Sections 11.1 – 11.2
Analytic Geometry and Circles
What Conic Sections Are · The Second-Degree Equation · Circles in Standard Form
Section 11.1

Analytic Geometry as Algebra and Shape

Analytic geometry is the branch of mathematics that uses a coordinate system to study geometric figures through their equations. It unites two great traditions: the geometry of shapes developed by the ancient Greeks, and the algebra of equations developed through the Renaissance and into the seventeenth century. The French mathematician René Descartes is usually credited with formalizing the connection between algebra and geometry. By assigning numerical coordinates to points in the plane, Descartes made it possible to describe curves as sets of points satisfying algebraic conditions.

A straight line is no longer just a ruler-drawn object; it is the set of all points (x, y) satisfying y = 2x + 3. A circle is the set of all points a fixed distance from a center, described by x² + y² = r². Conic sections are among the most important families of curves in analytic geometry.

What Is a Conic Section?

A double cone is a three-dimensional figure formed by two identical cones placed tip-to-tip at the origin, extending infinitely in both directions. When a flat plane intersects this double cone, the intersection traces a curve. Depending on the angle and position of the plane, that curve is one of four types.

Plane OrientationConic SectionDescription
Perpendicular to axisCircleEqual distance in all directions from the cone's axis
Tilted slightly from perpendicularEllipseCuts one nappe of the cone in a closed loop
Parallel to one side of the coneParabolaCuts one nappe, opening infinitely in one direction
Cuts both nappesHyperbolaTwo separate open branches, one on each nappe

The Role of the Second-Degree Equation

Every conic section is described algebraically by a second-degree equation in x and y. The general second-degree equation (with no xy cross-term) has the form:

General Second-Degree Equation
Ax² + Cy² + Dx + Ey + F = 0
When B = 0 (no xy cross-term), the conic's axes are aligned with the coordinate axes.
Identifying Conics from Coefficients
  • A = C (equal nonzero values, same sign) → Circle
  • A ≠ C, both same sign → Ellipse
  • Exactly one of A, C is zero → Parabola
  • A and C have opposite signs → Hyperbola
MGU Library Connection
  • Appendix M: Conic Section Formula Reference — visual summary of all four conic types and their equations
  • Algebra II, Chapter 10: Introduction to Conic Sections
Section 11.2

Circles

The circle is the most familiar of the four conics. A circle is defined as the set of all points in a plane at a fixed distance from a given center point. Let the center be (h, k) and the radius be r > 0. A point (x, y) lies on the circle if and only if its distance from the center equals r. Applying the distance formula and squaring gives the standard form.

Standard Form of a Circle
(x − h)² + (y − k)² = r²
Center: (h, k) · Radius: r · When center is at origin: x² + y² = r²
FeatureValue
Center(h, k)
Radiusr
Diameter2r
Domain[h − r, h + r]
Range[k − r, k + r]
SymmetrySymmetric about any line through the center
Example 11.2.1
Writing the equation of a circle
  • ProblemWrite the standard form equation of the circle with center (3, −2) and radius 5.
  • ThinkUse standard form (x − h)² + (y − k)² = r² with h = 3, k = −2, and r = 5.
  • RepresentStandard form of a circle.
  • Solve (x − 3)² + (y − (−2))² = 5² (x − 3)² + (y + 2)² = 25
  • CheckCheck the topmost point (3, 3): (3 − 3)² + (3 + 2)² = 0 + 25 = 25 ✓
Answer: (x − 3)² + (y + 2)² = 25
Example 11.2.2
Identifying center and radius
  • ProblemIdentify the center and radius of (x + 1)² + (y − 4)² = 9.
  • ThinkRewrite to match (x − h)² + (y − k)² = r² by reading h, k, and r² directly. Note (x + 1)² = (x − (−1))².
  • Solveh = −1, k = 4, r² = 9, so r = 3.
Answer: Center (−1, 4), radius 3.
Graph Instruction — Circle

Draw a circle centered at (−1, 4) with radius 3. Mark the center with a dot. Mark the four boundary points: directly above (−1, 7), below (−1, 1), left (−4, 4), and right (2, 4). Draw the circle through these four points. Label the center and the radius.

Student Warning — Reading the Center's Sign

In standard form (x − h)² + (y − k)² = r², the center uses subtraction inside the parentheses. When the equation shows (x + 1)², the center has x-coordinate −1, not +1. Always rewrite as (x − (−1))² to read the center correctly. This same issue affects ellipses and hyperbolas.

Practice 11.2Circles

Write equations or identify features as indicated.

  1. Center (0, 0), radius 6
  2. Center (−3, 5), radius √7
  3. Center (4, −1), diameter 10
  4. Identify center and radius: (x − 2)² + (y + 5)² = 49
  5. Identify center and radius: (x + 3)² + y² = 16
  6. Does (7, −2) lie on circle (x − 3)² + (y + 2)² = 25? Show work.
Section 11.3
Parabolas
Focus · Directrix · Four Orientations · Optics and Engineering Applications
Section 11.3

Parabolas

The parabola was encountered in Algebra I and Algebra II as the graph of a quadratic function. Here, the parabola is studied more deeply as a geometric curve defined by a focus-directrix relationship, and equations are written both in vertex form and standard conic form. Parabolas that open upward, downward, left, or right are all studied — including horizontal parabolas, which are not functions.

Geometric Definition of a Parabola

A parabola is the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix).

The vertex is the point on the parabola closest to the directrix and focus. The signed distance from the vertex to the focus is p. When p > 0 the parabola opens away from the directrix in the positive direction; when p < 0 it reverses.

Standard Forms — Vertex at (h, k)

OrientationStandard FormFocusDirectrixAxis
Vertical (up if p > 0, down if p < 0)(x − h)² = 4p(y − k)(h, k + p)y = k − px = h
Horizontal (right if p > 0, left if p < 0)(y − k)² = 4p(x − h)(h + p, k)x = h − py = k
Why the Focus-Directrix Definition?

A parabolic mirror or dish is designed so that incoming parallel rays all reflect to the same focus point. Conversely, a light source at the focus reflects off the parabola as perfectly parallel rays. This is why flashlights, car headlights, satellite dishes, and reflecting telescopes use parabolic shapes. The geometric equidistance property produces this exact focusing behavior.

Example 11.3.1
Writing the equation from focus and directrix
  • ProblemFind the equation of the parabola with vertex at the origin, axis along the y-axis, and focus at (0, 3).
  • ThinkThe focus is above the vertex, so the parabola opens upward. The form is x² = 4py with p = 3.
  • RepresentVertical parabola: x² = 4py.
  • Solvex² = 4(3)y = 12y
  • CheckFocus at (0, p) = (0, 3) ✓. Directrix: y = −3. Vertex (0, 0): distance to focus = 3, distance to directrix = 3 ✓.
Answer: x² = 12y
Example 11.3.2
Identifying features from equation — horizontal parabola
  • ProblemIdentify the vertex, focus, directrix, and direction of opening for (y + 2)² = −8(x − 1).
  • ThinkThe y variable is squared, so this is a horizontal parabola: (y − k)² = 4p(x − h). Read k = −2, h = 1, and 4p = −8, so p = −2.
  • Solve Vertex: (h, k) = (1, −2) Opens: left (p = −2 < 0) Focus: (h + p, k) = (1 + (−2), −2) = (−1, −2) Directrix: x = h − p = 1 − (−2) = 3
  • CheckVertex (1, −2): distance to focus = |1 − (−1)| = 2 ✓; distance to directrix = |1 − 3| = 2 ✓.
Answer: Vertex (1, −2); opens left; focus (−1, −2); directrix x = 3.
Graph Instruction — Horizontal Parabola

Draw a horizontal parabola opening to the left with vertex at (1, −2). Mark the focus at (−1, −2) with a dot and label it. Draw the directrix as a dashed vertical line at x = 3 and label it. Draw the axis of symmetry as a dashed horizontal line at y = −2. The parabola should open leftward, wider than a standard parabola since |4p| = 8.

Student Warning — Horizontal Parabolas Are Not Functions

A horizontal parabola (y² = ···) fails the vertical line test and is therefore not a function. Do not confuse it with the parabolas-as-functions from Algebra II. In that setting, only vertical parabolas (opening up or down) were treated as functions. In conic section study, all four directions are studied as curves without this restriction.

Practice 11.3Parabolas

Identify vertex, focus, and directrix, or write the equation as indicated.

  1. x² = 12y
  2. y² = −20x
  3. (x + 2)² = 8(y − 1)
  4. (y − 3)² = −4(x + 5)
  5. Vertex (0, 0), focus (0, −3)
  6. Vertex (2, 1), focus (5, 1)
  7. Vertex (−1, 4), directrix y = 6
  8. x² + 4x − 4y + 8 = 0 (complete the square first)
Section 11.4
Ellipses
Sum-of-Distances Definition · Semi-Major and Semi-Minor Axes · Foci · Orbits
Section 11.4

Ellipses

An ellipse is the set of all points in a plane for which the sum of the distances from two fixed points (the foci) is constant. Let the two foci be F₁ and F₂. Then a point P lies on the ellipse if and only if PF₁ + PF₂ = 2a, where 2a is the constant sum, always greater than the distance between the foci.

The ellipse has two special axes. The major axis passes through both foci; its half-length is a (the semi-major axis). The minor axis is perpendicular to the major axis at the center; its half-length is b (the semi-minor axis). The distance from the center to each focus is c, and these three values are related by:

Ellipse Focal Relationship
a² = b² + c²   or equivalently   c² = a² − b²
Always: a > b > 0 and a > c > 0. The foci lie inside the ellipse.

Standard Forms of Ellipses — Center at (h, k)

FormStandard EquationVertices (major axis)Foci
Horizontal major axis[(x−h)²/a²] + [(y−k)²/b²] = 1, a > b(h ± a, k)(h ± c, k)
Vertical major axis[(x−h)²/b²] + [(y−k)²/a²] = 1, a > b(h, k ± a)(h, k ± c)
Key Identification Rule for Ellipses

The larger denominator is always a². The variable under the larger denominator tells you the direction of the major axis: if a² is under x², the major axis is horizontal; if a² is under y², the major axis is vertical. Do not assume the major axis is always horizontal just because x appears first.

Example 11.4.1
Writing and graphing an ellipse
  • ProblemFind the center, vertices, co-vertices, and foci of (x−2)²/25 + (y+1)²/9 = 1.
  • ThinkThe larger denominator is 25, under (x−2)², so the major axis is horizontal. a² = 25, b² = 9, center at (2, −1).
  • Solve a = 5, b = 3 c² = 25 − 9 = 16, c = 4 Center: (2, −1) Vertices: (2 ± 5, −1) = (7, −1) and (−3, −1) Co-vertices: (2, −1 ± 3) = (2, 2) and (2, −4) Foci: (2 ± 4, −1) = (6, −1) and (−2, −1)
  • CheckVerify c < a: 4 < 5 ✓. Foci at x = −2 and x = 6 lie between vertices at x = −3 and x = 7 ✓.
Answer: Center (2, −1); vertices (7, −1) and (−3, −1); co-vertices (2, 2) and (2, −4); foci (6, −1) and (−2, −1).
Graph Instruction — Horizontal Ellipse

Draw an ellipse centered at (2, −1). Mark the two vertices at (−3, −1) and (7, −1) on the horizontal axis and label them. Mark the two co-vertices at (2, 2) and (2, −4) on the vertical axis and label them. Mark the two foci at (−2, −1) and (6, −1) as interior points (not on the ellipse) and label them. Draw the major axis as a dashed horizontal segment from vertex to vertex, and the minor axis as a dashed vertical segment from co-vertex to co-vertex. The ellipse is wider than it is tall.

Example 11.4.2
Writing an ellipse from given information
  • ProblemWrite the standard form equation of the ellipse with center (0, 0), one focus at (0, 4), and one vertex at (0, 5).
  • ThinkThe focus and vertex are both on the y-axis, so the major axis is vertical. Use x²/b² + y²/a² = 1 with a > b.
  • Solve a = 5 (vertex to center distance) c = 4 (focus to center distance) b² = a² − c² = 25 − 16 = 9 Equation: x²/9 + y²/25 = 1
Answer: x²/9 + y²/25 = 1
Student Warning — Confusing a and b

Students often assign a² to whichever denominator appears first, or assume the major axis is always horizontal. Remember: a² is always the larger denominator, and the direction of the major axis is determined by which variable sits under a². Always identify the larger denominator first before labeling anything as a, b, or c.

Practice 11.4Ellipses

Identify all key features, or write the equation from the given information.

  1. x²/36 + y²/16 = 1
  2. x²/4 + y²/25 = 1
  3. (x−1)²/49 + (y+2)²/9 = 1
  4. Center (0,0), vertex (7,0), co-vertex (0,4)
  5. Center (3,−1), vertex (3,4), one focus (3,2)
  6. Foci at (±5, 0), constant sum = 14
  7. 9x² + 16y² − 36x + 96y + 36 = 0 (complete the square)
Section 11.5
Hyperbolas
Difference-of-Distances Definition · Asymptotes · Two Branches · Transverse Axis
Section 11.5

Hyperbolas

A hyperbola is the set of all points in a plane for which the absolute difference of the distances from two fixed points (the foci) is constant: |PF₁ − PF₂| = 2a. This makes the hyperbola the "opposite" of an ellipse — an ellipse adds distances, a hyperbola subtracts them. A hyperbola has two separate branches, each curving away from the center. It has a transverse axis connecting the two vertices, a conjugate axis perpendicular to it at the center, and two asymptotes that the branches approach without ever reaching.

The relationship between a, b, and focal distance c is fundamentally different from the ellipse:

Hyperbola Focal Relationship
c² = a² + b²
Always: c > a. The foci lie outside the curve — beyond the vertices. Compare with the ellipse: c² = a² − b².

Standard Forms of Hyperbolas — Center at (h, k)

OrientationStandard FormVerticesAsymptotes
Horizontal transverse axis (opens left/right)[(x−h)²/a²] − [(y−k)²/b²] = 1(h ± a, k)y − k = ±(b/a)(x − h)
Vertical transverse axis (opens up/down)[(y−k)²/a²] − [(x−h)²/b²] = 1(h, k ± a)y − k = ±(a/b)(x − h)
The Asymptote Rectangle

Draw a rectangle centered at (h, k) with width 2a in the direction of the transverse axis and height 2b in the other direction. The asymptotes are the diagonals of this rectangle, extended as lines. Each branch of the hyperbola opens away from the center, bending toward one pair of corners of this rectangle. The branches get closer and closer to the asymptote lines but never touch them.

Example 11.5.1
Graphing a hyperbola — horizontal transverse axis
  • ProblemIdentify center, vertices, foci, and asymptotes of x²/16 − y²/9 = 1.
  • ThinkThe positive term is x², so the transverse axis is horizontal. Center at origin, a² = 16, b² = 9.
  • Solve a = 4, b = 3 c² = 16 + 9 = 25, c = 5 Center: (0, 0) Vertices: (±4, 0) Foci: (±5, 0) Asymptotes: y = ±(3/4)x
  • CheckFor a horizontal hyperbola, c > a: 5 > 4 ✓. Foci are outside the vertices along the x-axis ✓.
Answer: Center (0, 0); vertices (±4, 0); foci (±5, 0); asymptotes y = ±(3/4)x.
Graph Instruction — Horizontal Hyperbola

Draw a coordinate plane. Mark the center at the origin. Draw the asymptote rectangle: width 8 (from −4 to 4 along x-axis), height 6 (from −3 to 3 along y-axis), centered at the origin. Draw the diagonals of this rectangle as dashed lines — these are the asymptotes y = ±(3/4)x. Mark the vertices at (−4, 0) and (4, 0). Draw two branches: one opening right through (4, 0), one opening left through (−4, 0), each curving toward but never touching the asymptote lines. Mark the foci at (±5, 0) inside the plane as points not on the curve.

Example 11.5.2
Writing an equation from features — vertical hyperbola
  • ProblemWrite the standard form equation of the hyperbola with center (1, −3), vertices at (1, 0) and (1, −6), and foci at (1, 2) and (1, −8).
  • ThinkThe vertices and foci share the same x-coordinate as the center, so the transverse axis is vertical.
  • Solve a = |0 − (−3)| = 3 c = |2 − (−3)| = 5 b² = c² − a² = 25 − 9 = 16 Equation: (y + 3)²/9 − (x − 1)²/16 = 1
Answer: (y + 3)²/9 − (x − 1)²/16 = 1

Comparing Ellipse and Hyperbola

PropertyEllipseHyperbola
DefinitionPF₁ + PF₂ = 2a (sum)|PF₁ − PF₂| = 2a (difference)
Focal relationshipc² = a² − b²c² = a² + b²
Foci vs. verticesFoci inside the curve (c < a)Foci outside the curve (c > a)
ShapeOne closed curveTwo open branches
AsymptotesNoneTwo diagonal lines
Student Warning — Ellipse vs. Hyperbola Formula

For an ellipse: c² = a² − b² (foci are inside, so c < a, so c² < a², so you subtract).
For a hyperbola: c² = a² + b² (foci are outside, so c > a, so c² > a², so you add).
A common error is using the ellipse formula for a hyperbola, producing a negative c², which is impossible. Always check whether your answer has c > a (hyperbola) or c < a (ellipse).

Practice 11.5Hyperbolas

Identify all key features, or write the equation from the given information.

  1. x²/9 − y²/16 = 1
  2. y²/4 − x²/25 = 1
  3. (x+3)²/16 − (y−1)²/9 = 1
  4. Center (0,0), vertices (±3,0), foci (±5,0)
  5. Center (2,−4), vertex (2,−1), conjugate axis length 6
  6. Asymptotes y = ±(2/3)x, vertex at (3, 0)
  7. 4x² − y² − 8x + 6y − 21 = 0 (complete the square)
Sections 11.6 – 11.8
Completing the Square, Translation, and Identification
General Form to Standard Form · Translated Conics · Reading Equations at a Glance
Section 11.6

Completing the Square for Conics

Conic equations often appear in general form — expanded, with squared and linear terms mixed together. To analyze a conic's features, it must be converted to standard form. The tool for doing this is completing the square, extended here to two variables simultaneously.

General Procedure

1. Group x-terms and y-terms. Move the constant to the right side.

2. Factor out the leading coefficient of each squared term if not already 1.

3. Complete the square for the x-group and y-group separately. Add the same values to both sides.

4. Write each completed square as a binomial squared.

5. Divide both sides if needed to reach standard form.

Example 11.6.1
Completing the square — circle
  • ProblemWrite x² + y² − 6x + 4y − 3 = 0 in standard form.
  • ThinkBoth variables squared with equal coefficients: circle. Group and complete the square for each variable.
  • Solve x² − 6x + y² + 4y = 3 Complete x: (−6/2)² = 9. Add 9 to both sides. Complete y: (4/2)² = 4. Add 4 to both sides. (x² − 6x + 9) + (y² + 4y + 4) = 3 + 9 + 4 (x − 3)² + (y + 2)² = 16
Answer: Circle with center (3, −2) and radius 4.
Example 11.6.2
Completing the square — ellipse with leading coefficients
  • ProblemWrite 4x² + 9y² − 8x + 36y + 4 = 0 in standard form.
  • ThinkTwo squared terms, different positive coefficients: ellipse. Factor out coefficients before completing the square.
  • Solve 4(x² − 2x) + 9(y² + 4y) = −4 For x: (−2/2)² = 1. Add 4 · 1 = 4 to both sides. For y: (4/2)² = 4. Add 9 · 4 = 36 to both sides. 4(x² − 2x + 1) + 9(y² + 4y + 4) = −4 + 4 + 36 4(x − 1)² + 9(y + 2)² = 36 Divide by 36: (x − 1)²/9 + (y + 2)²/4 = 1
Answer: Ellipse with center (1, −2), a² = 9 (horizontal major axis), b² = 4.
Example 11.6.3
Completing the square — hyperbola
  • ProblemWrite x² − 4y² − 2x + 16y − 19 = 0 in standard form.
  • ThinkTwo squared terms with opposite signs: hyperbola.
  • Solve (x² − 2x) − 4(y² − 4y) = 19 For x: (−2/2)² = 1. Add 1 to both sides. For y (inside −4): (−4/2)² = 4. Add −4 · 4 = −16 to both sides. (x² − 2x + 1) − 4(y² − 4y + 4) = 19 + 1 − 16 (x − 1)² − 4(y − 2)² = 4 Divide by 4: (x − 1)²/4 − (y − 2)² = 1
Answer: Hyperbola with center (1, 2), a² = 4, b² = 1, horizontal transverse axis.
Student Warning — Multiplying When Completing the Square

When the equation has −4(y² − 4y) and you add 4 inside the parentheses, you must add −4 × 4 = −16 to the right side of the equation, not just 4. The factor outside the parentheses multiplies whatever is added inside. This is the most common completing-the-square error with conics.

Section 11.7

Translated Conics

Translation is a rigid motion that shifts a curve horizontally and vertically without changing its shape, orientation, or size. The algebraic effect is replacing x with (x − h) and y with (y − k). All features — center, vertices, co-vertices, foci, asymptotes — shift by the same vector (h, k) from their origin-centered positions.

Origin-Centered FormTranslated Form
x² + y² = r²(x−h)² + (y−k)² = r²
x² = 4py(x−h)² = 4p(y−k)
x²/a² + y²/b² = 1(x−h)²/a² + (y−k)²/b² = 1
x²/a² − y²/b² = 1(x−h)²/a² − (y−k)²/b² = 1
Section 11.8

Identifying Conics from Equations

Given a second-degree equation with no xy cross-term, the conic type can be identified before converting to standard form, simply by examining the structure of the squared terms.

SituationConic
Only one variable is squaredParabola
Both squared, same nonzero coefficient, same signCircle
Both squared, different coefficients, same signEllipse
Both squared, opposite signsHyperbola
Example 11.8.1
Identifying conics — four cases
  • (a)3x² + 3y² − 12x + 6y − 5 = 0 → Both squared, equal coefficients → Circle
  • (b)x² + 4y² − 2x + 16y − 19 = 0 → Both squared, different positive coefficients → Ellipse
  • (c)y² + 6x − 4y + 1 = 0 → Only y is squared → Parabola (opens left or right)
  • (d)x² − 9y² + 4x + 36y − 41 = 0 → Both squared, opposite signs → Hyperbola
Practice 11.6 – 11.8Completing the Square and Identification

Identify the conic type, then convert to standard form and state all key features.

  1. x² + y² − 10x + 4y + 13 = 0
  2. x² + 4y² − 6x − 8y − 3 = 0
  3. 9x² − 16y² + 18x + 64y − 199 = 0
  4. y² − 4x − 8y + 28 = 0
  5. 4x² + 9y² + 16x − 18y − 11 = 0
  6. 2x² + 2y² − 4x + 12y − 6 = 0
Sections 11.9 – 11.12
Eccentricity and Applications
Focus and Directrix · Eccentricity · Astronomy and Optics · Engineering and Design
Section 11.9

Focus and Directrix

For the parabola, the focus-directrix definition was established in Section 11.3: every point on the parabola is equidistant from the focus and the directrix. The latus rectum is the chord passing through the focus and perpendicular to the axis of symmetry. It describes how wide the conic is at the focus and is useful for sketching accurate graphs.

ConicLatus Rectum Length
Parabola x² = 4py|4p|
Ellipse x²/a² + y²/b² = 12b²/a (through each focus)
Hyperbola x²/a² − y²/b² = 12b²/a (through each focus)
Section 11.10

Eccentricity as an Introduction

Eccentricity is a single number that measures how much a conic deviates from being a circle. It is denoted by e (not to be confused with Euler's number e ≈ 2.718, which uses the same letter in a different context — always read from context which meaning applies).

Eccentricity
e = c/a
c = distance from center to focus · a = semi-major axis (ellipse/hyperbola) or vertex-to-focus distance (parabola)
ConicValue of cEccentricity e = c/aShape
Circlec = 0e = 0Perfectly round
Ellipse0 < c < a0 < e < 1Elongated; more elongated as e → 1
Parabolac = a (in a sense)e = 1Open, unbounded
Hyperbolac > ae > 1Two open branches; wider as e increases
Eccentricity in Astronomy

Earth's orbit is an ellipse with eccentricity approximately 0.017 — nearly circular. Mars has eccentricity approximately 0.093. Comets often have orbits with eccentricities close to 1. Some objects passing through the solar system have hyperbolic trajectories (e > 1) and never return. Eccentricity is the single number that completely characterizes the shape of any conic.

Section 11.11

Conics in Astronomy and Optics

Johannes Kepler showed in 1609 that the planets travel in elliptical orbits with the sun at one focus. Isaac Newton later proved theoretically that any two objects interacting through an inverse-square gravitational law must travel in paths that are conic sections — ellipses for closed orbits, parabolas for trajectories at exactly escape velocity, and hyperbolas for trajectories above escape velocity.

In an elliptical orbit with semi-major axis a and focal distance c, the perihelion (closest point to the sun) is at distance a − c, and the aphelion (farthest point) is at distance a + c. This is why planets speed up when closer to the sun and slow down when farther.

Example 11.11.1
Planetary orbit from perihelion and aphelion
  • ProblemAn elliptical orbit has perihelion 1.4 AU and aphelion 2.6 AU. Write the equation of the orbit and find its eccentricity.
  • ThinkPerihelion = a − c = 1.4 and aphelion = a + c = 2.6. Solve this system.
  • Solve Adding: 2a = 4.0, so a = 2.0 Subtracting: 2c = 1.2, so c = 0.6 b² = a² − c² = 4.0 − 0.36 = 3.64 Equation: x²/4 + y²/3.64 = 1 Eccentricity: e = c/a = 0.6/2.0 = 0.3
Answer: x²/4 + y²/3.64 = 1; eccentricity 0.3.
Example 11.11.2
Satellite dish focal point
  • ProblemA satellite dish is 3 meters wide and 1.2 meters deep at its center. Where should the receiver be placed?
  • ThinkSet up a vertical parabola with vertex at the origin. The rim is at x = ±1.5 (half the width) and y = 1.2 (the depth). Use x² = 4py.
  • Solve Substitute (1.5, 1.2): 1.5² = 4p(1.2) 2.25 = 4.8p p = 2.25/4.8 ≈ 0.47 meters
  • CheckThe focus at p = 0.47 m is inside the dish (0.47 < 1.2 m deep) ✓. Reasonable position for a receiver mount.
Answer: The receiver should be placed approximately 47 cm above the vertex (center bottom) of the dish.
Section 11.12

Conics in Engineering and Design

Conics appear throughout engineering, architecture, and design. An elliptical room (whispering gallery) has the remarkable acoustic property that a whisper at one focus can be heard clearly at the other focus, because all sound waves from one focus reflect off the elliptical wall and converge at the other focus. Famous examples include the National Statuary Hall in the United States Capitol.

Hyperbolic cooling towers at power plants use the hyperbola's inward curve for structural strength with minimal material and efficient air flow. Parabolic arch bridges use the parabola's load-distribution properties. The LORAN navigation system (used before GPS) located ships by measuring time differences between radio signals from two stations. Since a constant difference of distances from two fixed points traces a hyperbola, LORAN placed ships at the intersection of two hyperbolas.

Example 11.12.1
Whispering gallery — finding the foci
  • ProblemA whispering gallery has an elliptical cross-section 80 feet long and 34 feet wide. Where should two people stand to be at the foci?
  • ThinkMajor axis: 2a = 80, so a = 40. Minor axis: 2b = 34, so b = 17.
  • Solve c² = a² − b² = 1600 − 289 = 1311 c = √1311 ≈ 36.2 feet Distance from end wall: a − c = 40 − 36.2 = 3.8 feet
Answer: Each person should stand approximately 3.8 feet from the nearest end wall of the room, near each end of the long axis.
Practice 11.9 – 11.12Eccentricity and Applications
  1. Find the eccentricity of the ellipse x²/100 + y²/64 = 1.
  2. Find the eccentricity of the hyperbola x²/9 − y²/16 = 1.
  3. An ellipse has a = 10 and e = 0.6. Find c and b.
  4. A parabolic solar collector has cross-section x² = 16y (in inches). At what height above the bottom should the heating element be placed?
  5. A bridge arch is parabolic, 100 m wide at its base and 40 m high at center. Write the equation with vertex at the top.
  6. A comet has orbital equation x²/4 − y²/12 = 1 (in AU). Find the eccentricity and the closest approach to the sun (at one focus).
  7. LORAN stations are located at (±200, 0) miles. A ship determines a distance difference of 240 miles. Write the hyperbola equation for the ship's location.
Sections 11.13 – 11.14
Common Mistakes and the Road Ahead
Eight Recurring Errors · Parametric Preview · Bridge to Chapter 12
Section 11.13

Common Conic Section Mistakes

Mistake 1 — Reading the center's sign incorrectly
"Center of (x + 4)² + (y − 1)² = 9 is (4, 1)."

Correct: (x + 4)² = (x − (−4))², so the center is (−4, 1). Always rewrite to see the subtraction explicitly.

Mistake 2 — Confusing a and b in ellipses
"For x²/9 + y²/25 = 1, the major axis is horizontal because x appears first."

Correct: a² is the larger denominator. Here a² = 25 is under y², so the major axis is vertical. Always identify the larger denominator first.

Mistake 3 — Using c² = a² − b² for hyperbolas
Finding foci of a hyperbola using c² = a² − b².

Correct: For ellipses: c² = a² − b². For hyperbolas: c² = a² + b². Hyperbola foci are outside the curve (c > a), so you must add. Subtracting often gives a negative value, which signals the wrong formula.

Mistake 4 — Forgetting to multiply when completing the square
4(x² − 2x + 1) + 9(y² + 4y + 4) = −4 + 1 + 4 (adding 1 and 4 to the right side)

Correct: Adding 1 inside the factor 4( ) adds 4 × 1 = 4 to the right side. Adding 4 inside the factor 9( ) adds 9 × 4 = 36 to the right side. The right side should be −4 + 4 + 36 = 36.

Mistake 5 — Thinking asymptotes touch the hyperbola
"The hyperbola intersects its asymptotes at large x-values."

Correct: Asymptotes are lines that the branches approach but never reach. The branches get closer and closer to the asymptote lines as the values grow but do not intersect them at any finite point.

Mistake 6 — Confusing horizontal and vertical orientation
Assuming ellipse major axis is always horizontal, or assuming the first squared term determines direction.

Correct: For ellipses, the major axis direction corresponds to the variable with the larger denominator. For hyperbolas, the transverse axis direction corresponds to the variable with the positive sign.

Mistake 7 — Treating e in eccentricity as Euler's number
Using e ≈ 2.718 in an eccentricity calculation.

Correct: In the context of conic sections, e stands for eccentricity (the ratio c/a), not the base of the natural logarithm. Context determines which meaning applies.

Mistake 8 — Graphing without identifying key features first
Drawing a rough curve before computing vertices, foci, or asymptotes.

Correct: Always identify center (or vertex), a, b, c, and relevant features before sketching. Rushing to a sketch without key features produces incorrectly shaped or positioned curves that are useless for problem solving.

Section 11.14

Preparing for Parametric Equations

Chapter 12 will introduce parametric equations — a way of describing a curve by giving both x and y as functions of a third variable called the parameter (usually t). Circles and ellipses can be described parametrically in a natural way that connects directly to the trigonometric functions studied in Parts I and II of this book.

For a circle of radius r centered at the origin, as t sweeps from 0 to 2π, the point (r cos t, r sin t) traces the entire circle. The parameter t is the angle on the unit circle — exactly the connection built in Chapter 4.

ConicParametric FormParameter Range
Circle x² + y² = r²x(t) = r cos t, y(t) = r sin t0 ≤ t ≤ 2π
Ellipse x²/a² + y²/b² = 1x(t) = a cos t, y(t) = b sin t0 ≤ t ≤ 2π
Hyperbola x²/a² − y²/b² = 1x(t) = a sec t, y(t) = b tan tappropriate intervals
Connection to the Unit Circle

The parametric form of the ellipse, x(t) = a cos t and y(t) = b sin t, is a direct generalization of the unit circle. When a = b = 1, it reduces exactly to the unit circle from Chapter 4. The parameter t represents the angle, and the coordinates (a cos t, b sin t) stretch the unit circle by factor a horizontally and b vertically. This is a geometric transformation of exactly the type studied in Chapter 2.

MGU Library Connection
  • Chapter 12: Parametric Equations, Polar Coordinates, and Vectors — full development of parametric and polar conic forms
  • Appendix M: Conic Section Formula Reference — complete standard forms and feature summaries
  • Appendix F: Unit Circle Reference — connection from trigonometric functions to parametric circle
  • Calculus I, Chapter 4: Optimization — conic applications; Calculus II, Chapter 10: Polar forms of conics
Chapter 11 — Back Matter

Chapter Summary

Key Terms Review

analytic geometry
The use of algebra and coordinate systems to study geometric figures through their equations.
conic section
A curve formed by the intersection of a plane and a double cone; includes circles, parabolas, ellipses, and hyperbolas.
circle
The set of all points equidistant from a center (h, k). Standard form: (x−h)² + (y−k)² = r².
parabola
The set of all points equidistant from a focus and a directrix. Opens in one of four directions based on the sign of p.
vertex
The turning point of a parabola; the point closest to both the focus and the directrix.
focus
A fixed point involved in the definition of a parabola, ellipse, or hyperbola. Ellipses and hyperbolas have two foci.
directrix
A fixed line in the definition of a parabola; every point on the parabola is equidistant from the focus and directrix.
ellipse
The set of all points for which the sum of distances from two foci equals the constant 2a.
major axis
The longer axis of an ellipse, passing through both foci. Half-length = a (semi-major axis).
minor axis
The shorter axis of an ellipse, perpendicular to the major axis at the center. Half-length = b (semi-minor axis).
co-vertices
The endpoints of the minor axis of an ellipse.
hyperbola
The set of all points for which the absolute difference of distances from two foci equals the constant 2a. Has two branches.
transverse axis
The axis of a hyperbola connecting the two vertices, passing through both foci.
conjugate axis
The axis of a hyperbola perpendicular to the transverse axis at the center. Half-length = b.
asymptotes
Lines that the branches of a hyperbola approach but never reach. They are the diagonals of the central rectangle.
completing the square
An algebraic technique that rewrites a quadratic in x (or y) as (x − h)² plus a constant, converting general conic form to standard form.
eccentricity (e)
The ratio c/a measuring how far a conic deviates from circular. Circle: e = 0; ellipse: 0 < e < 1; parabola: e = 1; hyperbola: e > 1.
latus rectum
The chord of a conic passing through a focus and perpendicular to the major axis or axis of symmetry.
standard form
A conic equation written with the center or vertex explicit and all squared terms on the left with coefficient 1 or 1/a².
general form
A second-degree equation expanded with all terms on one side: Ax² + Cy² + Dx + Ey + F = 0.

Concept Review Questions

  1. What is a conic section, and how does it arise geometrically from a double cone?
  2. How do you recognize a circle, parabola, ellipse, and hyperbola from their equations, without converting to standard form?
  3. How does the sign of p determine the direction a parabola opens?
  4. What is the geometric relationship between any point on a parabola and the focus and directrix?
  5. What is the sum-of-distances property of an ellipse? How is it used to verify that a point is on a given ellipse?
  6. How is c² = a² − b² for an ellipse different from c² = a² + b² for a hyperbola? Why does each formula make geometric sense?
  7. How are the asymptotes of a hyperbola related to the asymptote rectangle?
  8. What does eccentricity measure? What eccentricity value corresponds to each of the four conics?
  9. How does the focus-reflection property of parabolas explain the design of satellite dishes and car headlights?
  10. How do parametric equations connect circles and ellipses to trigonometric functions?

Skill Practice

Part AConic Identification

Identify each conic section without converting to standard form.

  1. x² + y² − 4x + 6y − 3 = 0
  2. x² + 4y² − 2x + 16y + 1 = 0
  3. y² + 6x − 4y − 2 = 0
  4. x² − y² + 4x − 2y − 4 = 0
  5. 9x² + 9y² − 36 = 0
  6. 4x² − 16y² + 8x + 64y − 100 = 0
  7. x² + 2x − 8y + 17 = 0
  8. 3x² + 5y² − 12x + 10y − 7 = 0
Part BCircles

Write standard form and identify center and radius, or write the equation from given information.

  1. Center (0, 0), radius 6
  2. Center (−3, 5), radius √7
  3. Center (4, −1), diameter 10
  4. x² + y² − 8x + 2y − 8 = 0
  5. x² + y² + 6x − 10y + 22 = 0
  6. 2x² + 2y² − 4x + 12y − 6 = 0
Part CParabolas

Find vertex, focus, and directrix; or write the equation from given information.

  1. x² = 12y
  2. y² = −20x
  3. (x + 2)² = 8(y − 1)
  4. (y − 3)² = −4(x + 5)
  5. x² + 4x − 4y + 8 = 0
  6. y² − 6y − 8x + 1 = 0
  7. Vertex (0,0), focus (0, −3)
  8. Vertex (2,1), focus (5,1)
Part DEllipses

Identify all features, or write the equation from given information.

  1. x²/36 + y²/16 = 1
  2. x²/4 + y²/25 = 1
  3. (x−1)²/49 + (y+2)²/9 = 1
  4. 9x² + 16y² − 36x + 96y + 36 = 0
  5. Center (0,0), vertex (7,0), co-vertex (0,4)
  6. Foci (±5, 0), sum of distances = 14
Part EHyperbolas

Identify all features including asymptotes, or write the equation from given information.

  1. x²/9 − y²/16 = 1
  2. y²/4 − x²/25 = 1
  3. (x+3)²/16 − (y−1)²/9 = 1
  4. 4x² − y² − 8x + 6y − 21 = 0
  5. Center (0,0), vertices (±3,0), foci (±5,0)
  6. Asymptotes y = ±(2/3)x, vertex at (3, 0)
Part FEccentricity
  1. Find e for x²/100 + y²/64 = 1
  2. Find e for x²/9 − y²/16 = 1
  3. Ellipse: a = 10, e = 0.6. Find c and b.
  4. Conic: e = 1.5, a = 4. Find c and identify conic type.

Application Problems

Define variables, write equations, and interpret results in context.

  1. A parabolic solar collector has cross-section x² = 16y (inches). At what height above the vertex should the heating element be placed?
  2. A comet follows a hyperbolic path with the sun at one focus: x²/4 − y²/12 = 1 (in AU). Find eccentricity and distance of closest approach to the sun.
  3. A bridge arch is parabolic, 100 m wide at its base and 40 m high at center. With vertex at the top, write the parabola equation. How high is the arch 20 m from center?
  4. LORAN stations are at (±200, 0) miles. A ship determines a distance difference of 240 miles. Write the hyperbola equation for the ship's location.
  5. An elliptical orbit has perihelion 1.0 AU and aphelion 3.4 AU. Find the eccentricity and write the equation of the ellipse (center at origin).
  6. A whispering gallery is 100 feet long and 40 feet wide. How far from each end wall should speakers stand to be at the foci?

Error Analysis

Each problem shows a student's work. Identify the error and write the correct answer.

Problem 1
"Center of (x + 4)² + (y − 1)² = 9 is (4, 1)."

What is wrong? What is the correct center?

Problem 2
Completing the square: 4(x² − 2x + 1) + 9(y² + 4y + 4) = −4 + 1 + 4 = 1

Identify all errors. What are the correct values added to the right side?

Problem 3
"Asymptotes of y²/16 − x²/9 = 1 are y = ±(4/3)x."

Is this correct? If not, find the correct asymptotes. (Hint: for a vertical hyperbola, the asymptote slope uses a/b, not b/a.)

Problem 4
"For x²/9 + y²/25 = 1, the major axis is horizontal because x² appears first."

Identify and correct the error.

Problem 5
For hyperbola x²/16 − y²/9 = 1: c² = a² − b² = 16 − 9 = 7, so c = √7.

Identify the error and find the correct value of c.

Chapter 11 Checkpoint

Chapter 11 Checkpoint
Complete without assistance — assesses all key skills from Chapter 11
Part I — Short Answer
  1. Write the equation of the circle with center (−2, 5) and radius √11.
  2. Find the vertex, focus, and directrix of (y − 1)² = −12(x + 3).
  3. Write the standard form of the ellipse with foci at (0, ±3) and vertices at (0, ±5). Find b and write the equation.
  4. Find the asymptotes of the hyperbola (x − 2)²/9 − (y + 1)²/16 = 1.
  5. Convert x² + y² − 4x + 10y + 20 = 0 to standard form and identify the conic.
  6. Convert 9x² − 4y² − 36x + 8y − 4 = 0 to standard form. Identify all key features.
Part II — Graphing
  1. Graph (x+1)²/25 + (y−2)²/16 = 1. Label the center, vertices, co-vertices, and foci.
  2. Graph x²/9 − y²/4 = 1. Include and label both asymptotes. Mark vertices and foci.
Part III — Applications
  1. A parabolic satellite dish is 4 feet wide and 1 foot deep at its center. Where should the receiver be placed? Show all work.
  2. An asteroid follows an elliptical orbit. Its closest approach to the sun is 1.0 AU and its farthest distance is 3.4 AU. Find the eccentricity of the orbit and write the equation of the ellipse with center at the origin.
Part IV — Error Analysis
  1. A student finds the foci of an ellipse using c² = a² + b². What error did the student make? What is the correct formula, and when would c² = a² + b² be correct?
  2. A student says the asymptotes of x²/4 − y²/9 = 1 are y = ±(2/3)x. Is this correct? Verify or correct.

(Answers may be placed in the answer key.)

Looking Ahead — Bridge Note
Chapter 12: Parametric Equations, Polar Coordinates, and Vectors

Chapter 11 described curves using equations in x and y. Chapter 12 introduces two new languages for describing position, motion, and curves. Parametric equations give x and y each as functions of a parameter t, which is natural for motion along a path — a planet orbiting the sun, a projectile in flight, a point tracing a circle. Polar coordinates locate points using distance r and angle θ, which is natural for curves with rotational symmetry. Vectors describe quantities that have both magnitude and direction — force, velocity, displacement — and form the mathematical language of physics and engineering. All three topics in Chapter 12 grow naturally from the trigonometry, conics, and function work done so far in this book.

MGU Library Connection
  • Appendix M: Conic Section Formula Reference — full standard forms, key features, and relationships
  • Appendix B: Geometry and Right Triangle Review — foundational geometric ideas underlying conic definitions
  • Algebra II, Chapter 10: Introduction to Conic Sections — prior coverage of circles and parabolas
  • Chapter 12: Parametric Equations, Polar Coordinates, and Vectors — parametric and polar forms of conics
  • Calculus I, Chapter 4: Optimization — conic applications involving maxima and minima
  • Calculus II, Chapter 10: Parametric Equations and Polar Coordinates — arc length, area, and polar conic forms
  • MGU Physics Series — orbital mechanics, optics, and Kepler's laws