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GELE MathematicsAnalytic GeometryStudy Notes

Study notes for Analytic Geometry that match the GELE 2026 syllabus. Built to mirror how Professional Regulation Commission (PRC) — Board of Geodetic Engineering structures GELE Mathematics questions, these notes walk through each concept with examples, formulas, and practice questions designed for time-pressured exam conditions.

Exam context

The Geodetic Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Geodetic Engineering and is scheduled for September 2026. The Mathematics subtest is marked as "Core" in the official pattern, and Analytic Geometry appears in position 4th of 10 in the GELE Mathematics review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent GELE 2026 papers have drawn roughly a meaningful share of questions from this subject.

Analytic Geometry - Study Notes

Analytic geometry is the mathematical framework that connects algebraic equations with geometric shapes using coordinate systems. For civil engineers, this discipline is fundamental to surveying (coordinate systems and distance calculations), structural design (locating points and lines), highway and railway curve design (parabolic and circular alignments), and computer-aided design (CAD) applications. This chapter develops proficiency in distance calculations, line equations, and the classification of conic sections—all essential for the PRC Civil Engineer Licensure Examination. The coordinate system allows us to describe geometric objects (points, lines, circles, parabolas) as algebraic equations, bridging pure mathematics with practical engineering applications.

Summary

Analytic geometry bridges algebra and geometry, providing the mathematical tools essential for civil engineering practice. This comprehensive study notes covers: **Fundamental Tools:** - Distance and midpoint formulas for coordinate-based surveying - Slope calculations and line equations for alignment design and property boundaries - Point-to-line distance for clearance verification and offset calculations **Conic Sections and Engineering Applications:** 1. **Circle (e = 0):** Roundabouts, water tanks, circular curves, areas of influence 2. **Ellipse (0 < e < 1):** Transition curves, architectural features, structural analysis 3. **Parabola (e = 1):** Highway vertical alignment (critical per NSCP 2015), arch structures, drainage design 4. **Hyperbola (e > 1):** Navigation systems (LORAN), specialized structural forms **Key Formulas to Master:** - Distance: d = √[(Δx)² + (Δy)²] - Slope: m = Δy/Δx; perpendicular condition: m₁ × m₂ = −1 - Point-to-line distance: d = |Ax₀ + By₀ + C|/√(A² + B²) - Conic identification: Same signs & same coeff (circle) → Different coeff (ellipse) → Opposite signs (hyperbola) → One squared term (parabola) - Focal relationships: Ellipse c² = a² − b²; Hyperbola c² = a² + b² - Eccentricity: e = 0 (circle), 0 < e < 1 (ellipse), e = 1 (parabola), e > 1 (hyperbola) **Practical Competencies Developed:** After mastering this chapter, you will be able to: 1. Calculate distances and midpoints for surveying and layout tasks 2. Determine line equations for boundaries, alignments, and reference axes 3. Verify perpendicularity and parallelism for structural design 4. Calculate clearance distances from structures to property lines or reference lines 5. Design and analyze circular curves, elliptical features, and parabolic alignments 6. Convert between different forms of conic equations 7. Identify and classify conics from their equations or geometric properties 8. Locate foci and directrices for structural design considerations 9. Calculate eccentricity and use it to understand curve behavior 10. Apply these concepts to highway design (NSCP 2015), surveying, and structural layout **Board-Exam Focus Areas:** The PRC Civil Engineer Licensure Examination emphasizes: - Distance and midpoint calculations from coordinates - Line equation determination and slope relationships - Point-to-line distance for clearance and offset problems - Classification of conics and identification of key parameters - Parabolic curves for highway vertical alignment design - Conversion between conic equation forms - Calculation of foci, directrices, and eccentricity - Practical application to real-world civil engineering scenarios This chapter provides the mathematical foundation for succeeding in the examination and practicing civil engineering with confidence in coordinate-based design and surveying applications.

Sections

The foundation of analytic geometry is the placement of geometric objects on a coordinate plane. A point P(x, y) is defined by its horizontal (x) and vertical (y) coordinates. Understanding how to measure distances and locate midpoints between points is crucial for surveying applications and structural layout in civil engineering. **Distance Formula Derivation:** Consider two points A(x₁, y₁) and B(x₂, y₂). By the Pythagorean theorem, the distance d between them is: d = √[(x₂ - x₁)² + (y₂ - y₁)²] This formula is used constantly in surveying to calculate distances between survey stations, and in structural engineering to determine member lengths in coordinate-based design. **Midpoint Formula:** The midpoint M of a segment AB is: M = ((x₁ + x₂)/2, (y₁ + y₂)/2) This is useful in civil engineering for finding the center of a plot of land or the midpoint of a structural member when performing symmetry analyses. **Practical Engineering Application:** In land surveying, if a plot corner at (100, 200) m is to be connected to a corner at (400, 500) m (coordinates relative to a site datum), the distance calculation ensures accurate boundary marking. Similarly, when designing a symmetric building about a centerline, the midpoint formula locates the geometric center.

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1. Fundamental Concepts: Points and Distances

Examples

Problem

Find the distance and midpoint between points A(2, 3) and B(8, 11).

Solution

Step 1: Apply distance formula: d = √[(8 - 2)² + (11 - 3)²] d = √[6² + 8²] d = √[36 + 64] d = √100 d = 10 units Step 2: Apply midpoint formula: M = ((2 + 8)/2, (3 + 11)/2) M = (10/2, 14/2) M = (5, 7) Answer: Distance = 10 units; Midpoint = (5, 7)

Problem

In a rectangular survey, two reference points are at (150, 250) m and (550, 650) m relative to a datum. Calculate the straight-line distance between them and the coordinates of the midpoint.

Solution

Step 1: Calculate distance: d = √[(550 - 150)² + (650 - 250)²] d = √[400² + 400²] d = √[160000 + 160000] d = √320000 d = 400√2 d ≈ 565.69 m Step 2: Calculate midpoint: M = ((150 + 550)/2, (250 + 650)/2) M = (700/2, 900/2) M = (350, 450) m Answer: Distance ≈ 565.69 m; Midpoint = (350, 450) m

Problem

Three survey stations form a triangle: P(0, 0), Q(300, 0), and R(300, 400) (all in meters). Verify that this forms a right triangle using the distance formula.

Solution

Step 1: Calculate distance PQ: PQ = √[(300 - 0)² + (0 - 0)²] = √[90000] = 300 m Step 2: Calculate distance PR: PR = √[(300 - 0)² + (400 - 0)²] = √[90000 + 160000] = √250000 = 500 m Step 3: Calculate distance QR: QR = √[(300 - 300)² + (400 - 0)²] = √[160000] = 400 m Step 4: Verify Pythagorean theorem: PQ² + QR² = 300² + 400² = 90000 + 160000 = 250000 = 500² = PR² Conclusion: Yes, PQR is a right triangle with the right angle at Q.

Key Points

  • Distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
  • Midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2)
  • Distance is always non-negative; it represents a magnitude
  • Useful for surveying, boundary determination, and coordinate-based design
  • The distance formula is a direct application of the Pythagorean theorem in 2D space

In civil engineering, lines represent alignments—street centerlines, property boundaries, structural axes, and survey traverse lines. Understanding slope, parallel and perpendicular relationships, and the various forms of line equations is essential for layout and design. **Slope (Gradient):** The slope m of a line through points (x₁, y₁) and (x₂, y₂) is: m = (y₂ - y₁)/(x₂ - x₁) = Δy/Δx Slope represents the steepness of a line. In civil engineering, slope also refers to the vertical rise per horizontal unit distance (e.g., a 5% grade means 5 m rise per 100 m horizontal). **Relationship Between Slopes:** - Parallel lines have equal slopes: m₁ = m₂ - Perpendicular lines have slopes whose product is –1: m₁ × m₂ = –1 (negative reciprocal) These relationships are critical when designing perpendicular structures, setting up coordinate systems, or verifying that survey measurements are consistent. **Line Equations:** A line can be expressed in several forms: 1. Point-Slope Form: y – y₁ = m(x – x₁) (Used when you know a point and slope) 2. Slope-Intercept Form: y = mx + b (Used when you know slope and y-intercept) 3. General Form: Ax + By + C = 0 (Standard form used in distance and intersection calculations) 4. Two-Point Form: (y – y₁)/(y₂ – y₁) = (x – x₁)/(x₂ – x₁) (Used when you know two points) The general form Ax + By + C = 0 is particularly useful because it applies to all lines (including vertical lines, which have undefined slope in slope-intercept form).

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2. Lines and Linear Equations

Examples

Problem

Find the equation of the line passing through points A(2, 3) and B(5, 9). Express in general form Ax + By + C = 0.

Solution

Step 1: Calculate slope: m = (9 - 3)/(5 - 2) = 6/3 = 2 Step 2: Use point-slope form with point A(2, 3): y - 3 = 2(x - 2) y - 3 = 2x - 4 y = 2x - 1 Step 3: Convert to general form: 2x - y - 1 = 0 Verification with point B(5, 9): 2(5) - 9 - 1 = 10 - 9 - 1 = 0 ✓ Answer: 2x - y - 1 = 0

Problem

A property line runs from corner P(100, 200) to corner Q(400, 200) (in meters). Find: (a) the equation of the property line, and (b) the equation of a perpendicular line through P.

Solution

Part (a): Equation of property line Step 1: Calculate slope: m₁ = (200 - 200)/(400 - 100) = 0/300 = 0 Step 2: Since slope is 0, this is a horizontal line: y = 200 Or in general form: 0x + 1y - 200 = 0, or y - 200 = 0 Part (b): Perpendicular line through P(100, 200) Step 1: For perpendicular slope: m₁ × m₂ = -1 0 × m₂ = -1 (This indicates perpendicular to horizontal is vertical) Step 2: A line perpendicular to horizontal is vertical: x = 100 Or in general form: x - 100 = 0 Answer: (a) Property line: y = 200; (b) Perpendicular line: x = 100

Problem

A road centerline has the equation 3x + 4y - 120 = 0. Find the slope of the road and the slope of a perpendicular line (such as a cross-drainage pipe).

Solution

Step 1: Convert general form to slope-intercept form: 3x + 4y - 120 = 0 4y = -3x + 120 y = (-3/4)x + 30 Slope of road: m₁ = -3/4 Step 2: Slope of perpendicular line: m₂ = -1/m₁ = -1/(-3/4) = 4/3 Verification: m₁ × m₂ = (-3/4) × (4/3) = -12/12 = -1 ✓ Answer: Road slope = -3/4; Perpendicular slope = 4/3

Problem

A surveyor measures a slope of 8% on a site access road. If the horizontal distance is 250 m, what is the vertical rise? Express the slope as m = Δy/Δx and in general form.

Solution

Step 1: Understand slope percentage: 8% slope = 8 m rise per 100 m horizontal = 0.08 So m = 0.08 Step 2: Calculate vertical rise: m = Δy/Δx 0.08 = Δy/250 Δy = 0.08 × 250 = 20 m Step 3: If the road starts at (0, 0): y = 0.08x (slope-intercept form) Step 4: Convert to general form: y = 0.08x 100y = 8x (multiply by 100 to remove decimals) 8x - 100y = 0 Or simplified: 2x - 25y = 0 Answer: Vertical rise = 20 m; General form: 8x - 100y = 0 (or 2x - 25y = 0)

Key Points

  • Slope m = Δy/Δx represents steepness and is the ratio of vertical to horizontal change
  • Parallel lines: m₁ = m₂
  • Perpendicular lines: m₁ × m₂ = –1
  • Point-slope form: y – y₁ = m(x – x₁)
  • Slope-intercept form: y = mx + b (b is y-intercept)
  • General form: Ax + By + C = 0 (works for all lines, including vertical)
  • To convert between forms: expand or rearrange algebraically
  • A line with slope m = 0 is horizontal (y = constant)
  • A vertical line has undefined slope (x = constant)

In civil engineering, calculating the perpendicular distance from a point to a line is essential for: - Determining the shortest distance from a structure to a property boundary - Checking clearance distances from a survey line - Finding the offset distance in road design - Calculating the perpendicular distance from a column to a reference axis **Point-to-Line Distance Formula:** The perpendicular distance d from a point P(x₀, y₀) to a line Ax + By + C = 0 is: d = |Ax₀ + By₀ + C| / √(A² + B²) The absolute value ensures distance is always positive. The denominator √(A² + B²) is the normalization factor that accounts for the scaling of coefficients A and B. **Derivation Insight:** The distance formula is derived from the fact that the perpendicular from a point to a line is the shortest path. The numerator |Ax₀ + By₀ + C| represents the signed distance (before normalization), and the denominator normalizes it to the actual perpendicular distance. **Important Note:** When using this formula, the line must be in general form Ax + By + C = 0. If given in another form, convert it first.

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3. Distance from a Point to a Line

Examples

Problem

Find the perpendicular distance from point P(3, 4) to the line 3x + 4y - 10 = 0.

Solution

Step 1: Identify coefficients from line equation 3x + 4y - 10 = 0: A = 3, B = 4, C = -10 Point: (x₀, y₀) = (3, 4) Step 2: Apply the point-to-line distance formula: d = |Ax₀ + By₀ + C| / √(A² + B²) d = |3(3) + 4(4) - 10| / √(3² + 4²) d = |9 + 16 - 10| / √(9 + 16) d = |15| / √25 d = 15 / 5 d = 3 units Answer: The perpendicular distance is 3 units.

Problem

A property boundary is defined by the line 2x + 3y - 18 = 0. A building corner is located at (6, 2). Calculate the perpendicular distance from the corner to the boundary. (All coordinates in meters.)

Solution

Step 1: Identify coefficients: Line: 2x + 3y - 18 = 0 A = 2, B = 3, C = -18 Point: (6, 2) Step 2: Check if point is on the line: 2(6) + 3(2) - 18 = 12 + 6 - 18 = 0 The point (6, 2) lies exactly on the boundary line, so distance = 0. Step 3: If we need to move 5 m perpendicular from this point: A unit vector perpendicular to the line is: n̂ = (A, B)/√(A² + B²) = (2, 3)/√13 New point = (6, 2) + 5 × (2/√13, 3/√13) = (6 + 10/√13, 2 + 15/√13) Answer: Distance = 0 m (corner is on the boundary)

Problem

A surveyed property line is described by the equation 5x - 12y + 65 = 0. An existing building occupies point (2, 3). How far is the building from the property line? (Express in meters.)

Solution

Step 1: Identify coefficients: Line: 5x - 12y + 65 = 0 A = 5, B = -12, C = 65 Point: (2, 3) Step 2: Apply distance formula: d = |5(2) - 12(3) + 65| / √(5² + (-12)²) d = |10 - 36 + 65| / √(25 + 144) d = |39| / √169 d = 39 / 13 d = 3 m Answer: The building is 3 m from the property line.

Problem

Two parallel survey traverse lines are: L₁: 3x + 4y - 20 = 0 and L₂: 3x + 4y + 30 = 0. Find the perpendicular distance between these parallel lines.

Solution

Step 1: For parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0: Distance = |C₁ - C₂| / √(A² + B²) Step 2: Identify coefficients: A = 3, B = 4, C₁ = -20, C₂ = 30 Step 3: Calculate distance: d = |-20 - 30| / √(3² + 4²) d = |-50| / √(9 + 16) d = 50 / √25 d = 50 / 5 d = 10 units Answer: The perpendicular distance between the parallel lines is 10 units. Note: This method works for any two parallel lines. Simply pick a point on one line and calculate its distance to the other line—you'll get the same result.

Key Points

  • Point-to-line distance: d = |Ax₀ + By₀ + C| / √(A² + B²)
  • Line must be in general form Ax + By + C = 0
  • Distance is always non-negative (use absolute value)
  • The denominator √(A² + B²) is the normalization factor
  • This is the perpendicular (shortest) distance from the point to the line
  • Critical for checking clearances and offsets in civil engineering
  • If the point lies on the line, the distance is zero

The circle is the simplest conic section and is fundamental in civil engineering for: - Determining circular curves in highway and railway design (NSCP 2015) - Designing roundabouts and traffic circles - Finding areas of influence around structures - CAD applications and geometric layouts **Standard Form of a Circle:** (x - h)² + (y - k)² = r² Where: - (h, k) = center of the circle - r = radius - The circle consists of all points equidistant (distance r) from the center **General Form of a Circle:** Expanding the standard form gives: x² + y² + Dx + Ey + F = 0 Where D = -2h, E = -2k, F = h² + k² - r² **Converting From General to Standard Form:** Complete the square for both x and y terms: 1. Group x and y terms: (x² + Dx) + (y² + Ey) + F = 0 2. Complete the square: (x + D/2)² - (D/2)² + (y + E/2)² - (E/2)² + F = 0 3. Simplify: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F Compare with (x - h)² + (y - k)² = r² to identify center and radius. **Key Properties:** - A circle passes through any point (x, y) satisfying the equation - The distance from the center to any point on the circle is always r - A circle has no eccentricity (eccentricity e = 0) - Useful for determining catchment areas, sight distances, and clearance zones

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4. Conic Sections: Circle

Examples

Problem

Write the equation of a circle with center (3, -2) and radius 5. Express in both standard and general forms.

Solution

Standard Form: Step 1: Apply (x - h)² + (y - k)² = r² Center: (h, k) = (3, -2) Radius: r = 5 (x - 3)² + (y - (-2))² = 5² (x - 3)² + (y + 2)² = 25 General Form: Step 2: Expand the standard form: (x - 3)² + (y + 2)² = 25 x² - 6x + 9 + y² + 4y + 4 = 25 x² + y² - 6x + 4y + 9 + 4 - 25 = 0 x² + y² - 6x + 4y - 12 = 0 Answer: Standard form: (x - 3)² + (y + 2)² = 25 General form: x² + y² - 6x + 4y - 12 = 0

Problem

Find the center and radius of the circle given by the equation x² + y² + 8x - 6y + 16 = 0.

Solution

Step 1: Rearrange: x² + 8x + y² - 6y + 16 = 0 Step 2: Complete the square for x terms: x² + 8x = (x + 4)² - 16 Step 3: Complete the square for y terms: y² - 6y = (y - 3)² - 9 Step 4: Substitute back: (x + 4)² - 16 + (y - 3)² - 9 + 16 = 0 (x + 4)² + (y - 3)² - 16 - 9 + 16 = 0 (x + 4)² + (y - 3)² - 9 = 0 (x + 4)² + (y - 3)² = 9 Step 5: Compare with (x - h)² + (y - k)² = r²: Center: (h, k) = (-4, 3) Radius: r = √9 = 3 Answer: Center = (-4, 3); Radius = 3 units

Problem

A circular traffic roundabout is designed with center at (150, 200) m and passes through the point (158, 206) m. Find the radius and write the equation of the roundabout boundary.

Solution

Step 1: Calculate radius using distance formula: Center C = (150, 200) Point P = (158, 206) r = √[(158 - 150)² + (206 - 200)²] r = √[8² + 6²] r = √[64 + 36] r = √100 r = 10 m Step 2: Write the equation (standard form): (x - 150)² + (y - 200)² = 10² (x - 150)² + (y - 200)² = 100 Step 3: Convert to general form (if needed): x² - 300x + 22500 + y² - 400y + 40000 = 100 x² + y² - 300x - 400y + 62400 = 0 Answer: Radius = 10 m Equation: (x - 150)² + (y - 200)² = 100

Problem

Determine if the point (2, 1) lies on, inside, or outside the circle (x + 1)² + (y - 2)² = 9.

Solution

Step 1: Identify circle parameters: Center: (-1, 2) Radius: r = 3 Step 2: Calculate distance from point (2, 1) to center (-1, 2): d = √[(2 - (-1))² + (1 - 2)²] d = √[3² + (-1)²] d = √[9 + 1] d = √10 ≈ 3.162 units Step 3: Compare distance to radius: d = √10 ≈ 3.162 > r = 3 Answer: The point (2, 1) is outside the circle because the distance from the point to the center is greater than the radius.

Key Points

  • Standard form: (x - h)² + (y - k)² = r²
  • General form: x² + y² + Dx + Ey + F = 0
  • Center (h, k) and radius r uniquely define a circle
  • Eccentricity of a circle is 0
  • Convert between forms by completing the square
  • Useful for highway curves, roundabouts, and area calculations
  • Distance from center to circumference is constant (= r)
  • For a point on the circle, the distance equation is satisfied with equality

The ellipse appears in civil engineering for: - Highway transition curves (spiral to circular curve connections) - Building floor plans (elliptical arches and domes) - Structural analysis involving eccentric loading - Surveying and coordinate transformations **Standard Form of an Ellipse (Horizontal Major Axis):** (x - h)²/a² + (y - k)²/b² = 1 Where: - (h, k) = center of the ellipse - a = semi-major axis (longer radius, along x-axis) - b = semi-minor axis (shorter radius, along y-axis) - a > b (for horizontal orientation) **For Vertical Major Axis:** (x - h)²/b² + (y - k)²/a² = 1 (Note: a and b swap positions; a > b still) **Key Relationships:** - Foci are located at distance c from the center, where c² = a² - b² - Eccentricity: e = c/a (always 0 < e < 1 for an ellipse) - The sum of distances from any point on the ellipse to the two foci is constant = 2a - When a = b, the ellipse becomes a circle (special case, e = 0) **Geometric Properties:** - An ellipse is symmetric about both its axes - The major axis passes through both foci - The minor axis is perpendicular to the major axis - Ellipses are used in highway design as transition curves (though parabolas are more common in modern NSCP 2015 designs)

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5. Conic Sections: Ellipse

Examples

Problem

Write the equation of an ellipse with center (2, 3), semi-major axis a = 5 (horizontal), and semi-minor axis b = 3.

Solution

Step 1: Apply the standard form (horizontal major axis): (x - h)²/a² + (y - k)²/b² = 1 Step 2: Substitute values: Center: (h, k) = (2, 3) a = 5, b = 3 (x - 2)²/25 + (y - 3)²/9 = 1 Step 3: Verify it's an ellipse (a ≠ b and a > b): a = 5 > b = 3 ✓ Answer: (x - 2)²/25 + (y - 3)²/9 = 1

Problem

Find the foci and eccentricity of the ellipse x²/25 + y²/9 = 1.

Solution

Step 1: Identify parameters: (x - 0)²/25 + (y - 0)²/9 = 1 Center: (0, 0) a² = 25 → a = 5 (semi-major axis, horizontal) b² = 9 → b = 3 (semi-minor axis) Step 2: Calculate c (distance from center to focus): c² = a² - b² = 25 - 9 = 16 c = 4 Step 3: Locate foci: Since major axis is horizontal, foci are at (±c, 0): Focus 1: (-4, 0) Focus 2: (4, 0) Step 4: Calculate eccentricity: e = c/a = 4/5 = 0.8 Answer: Foci: (-4, 0) and (4, 0); Eccentricity: e = 0.8

Problem

An elliptical building footprint has the equation (x - 100)²/400 + (y - 150)²/225 = 1 (coordinates in meters). Find the center, dimensions of the major and minor axes, and the foci location.

Solution

Step 1: Identify center: (x - 100)²/400 + (y - 150)²/225 = 1 Center: (h, k) = (100, 150) m Step 2: Identify semi-axes: a² = 400 → a = 20 m (horizontal, since 400 > 225) b² = 225 → b = 15 m (vertical) Step 3: Full axis lengths: Major axis length = 2a = 40 m Minor axis length = 2b = 30 m Step 4: Calculate focal distance: c² = a² - b² = 400 - 225 = 175 c = √175 = 5√7 ≈ 13.23 m Step 5: Locate foci (along major/horizontal axis): Focus 1: (100 - 13.23, 150) ≈ (86.77, 150) m Focus 2: (100 + 13.23, 150) ≈ (113.23, 150) m Step 6: Calculate eccentricity: e = c/a = 5√7/20 = √7/4 ≈ 0.661 Answer: Center = (100, 150) m Major axis = 40 m (horizontal) Minor axis = 30 m (vertical) Foci ≈ (86.77, 150) and (113.23, 150) m Eccentricity ≈ 0.661

Problem

Given the ellipse with equation (x - 1)²/16 + (y + 2)²/25 = 1, determine if the major axis is horizontal or vertical, and find all key parameters.

Solution

Step 1: Identify form: (x - 1)²/16 + (y - (-2))²/25 = 1 Center: (1, -2) Step 2: Compare denominators: Under x: 16 Under y: 25 Since 25 > 16, the larger denominator is with y, so major axis is VERTICAL. Step 3: Assign semi-axes: a = 5 (semi-major axis, vertical) b = 4 (semi-minor axis, horizontal) Step 4: Calculate c: c² = a² - b² = 25 - 16 = 9 c = 3 Step 5: Locate foci (along vertical axis through center): Focus 1: (1, -2 + 3) = (1, 1) Focus 2: (1, -2 - 3) = (1, -5) Step 6: Eccentricity: e = c/a = 3/5 = 0.6 Answer: Major axis is VERTICAL Center = (1, -2) Semi-major axis a = 5 (vertical) Semi-minor axis b = 4 (horizontal) Foci: (1, 1) and (1, -5) Eccentricity: e = 0.6

Key Points

  • Standard form: (x - h)²/a² + (y - k)²/b² = 1 (a > b for horizontal major axis)
  • Center: (h, k)
  • Semi-major axis: a; Semi-minor axis: b
  • Foci: c = √(a² - b²) from center along major axis
  • Eccentricity: e = c/a (0 < e < 1 for ellipse)
  • Sum of distances from any point to both foci = 2a
  • An ellipse is more elongated as eccentricity increases toward 1
  • When a = b, the ellipse is a circle (e = 0)
  • Useful for transition curves and architectural designs

The parabola is the most important conic in civil engineering applications: - Highway vertical alignment curves (profile grades, NSCP 2015) - Arch structures and suspension bridges (cables follow parabolic paths) - Drainage swales and surface water flow paths - Satellite dish and reflector antennas - Trajectory of projectiles in ballistics **Standard Forms of a Parabola:** Vertex at (h, k), opening to the right: (y - k)² = 4a(x - h) Focus: (h + a, k) Directrix: x = h - a Vertex at (h, k), opening upward: (x - h)² = 4a(y - k) Focus: (h, k + a) Directrix: y = k - a Where: - a = focal distance (distance from vertex to focus) - The parabola opens toward the focus - The directrix is on the opposite side of the vertex from the focus - Distance from any point on the parabola to focus = distance to directrix (defining property) **Key Properties:** - Vertex (h, k) is the turning point - Axis of symmetry passes through the vertex and focus - Eccentricity e = 1 (all parabolas have the same eccentricity) - If a > 0, parabola opens right (or up); if a < 0, opens left (or down) - The focal parameter 4a is the coefficient in the equation **Highway Vertical Alignment (NSCP 2015):** Vertical parabolic curves are standard for grade transitions. The equation (x - h)² = 4a(y - k) describes a vertical alignment where: - PVC (Point of Vertical Curve) = vertex - x represents station distance along the horizontal - y represents elevation - The parameter a determines curve length

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6. Conic Sections: Parabola

Examples

Problem

Write the equation of a parabola with vertex at (3, 2) opening to the right with focal distance a = 2.

Solution

Step 1: Apply standard form (opens right): (y - k)² = 4a(x - h) Step 2: Substitute values: Vertex: (h, k) = (3, 2) a = 2 (y - 2)² = 4(2)(x - 3) (y - 2)² = 8(x - 3) Step 3: Identify focus and directrix: Focus: (h + a, k) = (3 + 2, 2) = (5, 2) Directrix: x = h - a = 3 - 2 = 1 Answer: Equation: (y - 2)² = 8(x - 3) Focus: (5, 2) Directrix: x = 1

Problem

Find the vertex, focus, and directrix of the parabola (x - 1)² = -12(y + 3).

Solution

Step 1: Rewrite in standard form: (x - 1)² = -12(y - (-3)) Step 2: Identify parameters: Standard form: (x - h)² = 4a(y - k) Vertex: (h, k) = (1, -3) 4a = -12 → a = -3 Step 3: Since a < 0, parabola opens DOWNWARD. Step 4: Find focus (distance a below vertex): Focus: (h, k + a) = (1, -3 + (-3)) = (1, -6) Step 5: Find directrix (distance a above vertex): Directrix: y = k - a = -3 - (-3) = 0, so y = 0 Answer: Vertex = (1, -3) Focus = (1, -6) Directrix = y = 0 Opening: Downward

Problem

A vertical highway grade transition uses a parabolic curve with equation (x)² = 40(y), where x is horizontal station (in 100 m units) and y is elevation change (in meters). Find the vertex, focus, focal distance, and the elevation at station x = 2.

Solution

Step 1: Rewrite in standard form: x² = 40y (x - 0)² = 40(y - 0) Step 2: Identify parameters: Vertex: (h, k) = (0, 0) 4a = 40 → a = 10 m Step 3: Find focus and directrix: Focus: (0, 10) m Directrix: y = -10 m Step 4: Physical interpretation: At x = 0 (station 0): y = 0 (elevation reference) Vertical distance from vertex to focus: 10 m (focal distance = 10 m) Step 5: Find elevation at x = 2 (station 200 m): x² = 40y (2)² = 40y 4 = 40y y = 0.1 m = 100 mm elevation rise at station 200 m Answer: Vertex = (0, 0) Focus = (0, 10) Focal distance = 10 m Elevation at station 2 units: y = 0.1 m

Problem

A parabolic arch spans 20 m horizontally with a maximum height of 8 m at the center. Write the equation of the parabola using the vertex as origin, and find the height at 5 m from the center.

Solution

Step 1: Set up coordinate system with vertex (top of arch) at origin: Vertex: (0, 0) Arch endpoints: (-10, -8) and (10, -8) Step 2: Use standard form (opens downward): (x)² = 4a(y) Since opens downward, a < 0: (x)² = 4a(y) Step 3: Substitute endpoint (10, -8): (10)² = 4a(-8) 100 = -32a a = -100/32 = -25/8 Equation: x² = 4(-25/8)y = -12.5y Or: x² = -25y/2 Or: 2x² + 25y = 0 Step 4: Find height at x = 5 m: (5)² = -12.5y 25 = -12.5y y = -2 m Height from arch endpoints = |-2| = 2 m, so the point is 8 - 2 = 6 m above the endpoints, or 2 m below the vertex. Answer: Equation: x² = -12.5y (or 2x² + 25y = 0) Height at x = 5 m: 6 m above the endpoints (or 2 m below the peak)

Key Points

  • Standard form (opens right): (y - k)² = 4a(x - h)
  • Standard form (opens up): (x - h)² = 4a(y - k)
  • Vertex: (h, k)
  • Focus: distance a from vertex in direction of opening
  • Directrix: line distance a on opposite side of vertex from focus
  • Eccentricity: e = 1 (characteristic of all parabolas)
  • Defining property: distance to focus = distance to directrix for any point on parabola
  • Coefficient 4a determines the 'focal distance' parameter
  • Essential for highway vertical curves (NSCP 2015) and arch structures
  • Opens toward the focus; directrix is on the opposite side

The hyperbola has specialized applications in civil engineering: - Radar and navigation systems (LORAN positioning—two transmitters locate by hyperbolic curves) - Trajectory analysis under certain field conditions - Acoustic and electromagnetic wave propagation - Cooling towers (hyperboloid surfaces) **Standard Forms of a Hyperbola:** Opening Left-Right (Horizontal Transverse Axis): (x - h)²/a² - (y - k)²/b² = 1 Foci: (h ± c, k), where c² = a² + b² Asymptotes: y - k = ±(b/a)(x - h) Opening Up-Down (Vertical Transverse Axis): (y - k)²/a² - (x - h)²/b² = 1 Foci: (h, k ± c), where c² = a² + b² Asymptotes: y - k = ±(a/b)(x - h) Where: - (h, k) = center - a = distance from center to vertex along transverse axis - b = distance from center to point on asymptote perpendicular to transverse axis - c = distance from center to focus (always c > a for hyperbola) - The hyperbola has two separate branches - Vertices are at distance a from the center **Key Properties:** - Eccentricity: e = c/a > 1 (always greater than 1 for hyperbola) - The difference of distances from any point to the two foci is constant = 2a - Asymptotes define the "envelope" of the hyperbola as it extends to infinity - Unlike the ellipse (+), the hyperbola uses subtraction (−) - As e → 1⁺, the hyperbola branches become narrow; as e increases, they widen

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7. Conic Sections: Hyperbola

Examples

Problem

Write the equation of a hyperbola with center (0, 0), opening left-right, with a = 3 and b = 4.

Solution

Step 1: Apply standard form (horizontal transverse axis): (x - h)²/a² - (y - k)²/b² = 1 Step 2: Substitute values: Center: (h, k) = (0, 0) a = 3, b = 4 (x)²/9 - (y)²/16 = 1 or x²/9 - y²/16 = 1 Step 3: Calculate c: c² = a² + b² = 9 + 16 = 25 c = 5 Step 4: Identify key features: Vertices: (±3, 0) → (-3, 0) and (3, 0) Foci: (±5, 0) → (-5, 0) and (5, 0) Asymptotes: y = ±(4/3)x Eccentricity: e = c/a = 5/3 ≈ 1.667 Answer: x²/9 - y²/16 = 1

Problem

Find the vertices, foci, asymptotes, and eccentricity of the hyperbola (y - 2)²/16 - (x + 1)²/9 = 1.

Solution

Step 1: Identify parameters: (y - 2)²/16 - (x - (-1))²/9 = 1 Center: (h, k) = (-1, 2) a² = 16 → a = 4 (semi-transverse axis, vertical) b² = 9 → b = 3 (semi-conjugate axis) Step 2: Transverse axis orientation: Since y² term is positive, opens UP-DOWN (vertical transverse axis). Step 3: Calculate c: c² = a² + b² = 16 + 9 = 25 c = 5 Step 4: Vertices (distance a vertically from center): Vertex 1: (-1, 2 + 4) = (-1, 6) Vertex 2: (-1, 2 - 4) = (-1, -2) Step 5: Foci (distance c vertically from center): Focus 1: (-1, 2 + 5) = (-1, 7) Focus 2: (-1, 2 - 5) = (-1, -3) Step 6: Asymptotes (vertical hyperbola formula): y - k = ±(a/b)(x - h) y - 2 = ±(4/3)(x - (-1)) y - 2 = ±(4/3)(x + 1) y = 2 ± (4/3)(x + 1) Asymptote 1: y = 2 + (4/3)(x + 1) → 4x - 3y + 10 = 0 Asymptote 2: y = 2 - (4/3)(x + 1) → 4x + 3y - 2 = 0 Step 7: Eccentricity: e = c/a = 5/4 = 1.25 Answer: Vertices: (-1, 6) and (-1, -2) Foci: (-1, 7) and (-1, -3) Asymptotes: 4x - 3y + 10 = 0 and 4x + 3y - 2 = 0 Eccentricity: e = 1.25

Problem

A LORAN navigation system uses two transmitters at (-100, 0) and (100, 0) (in km). A ship receives a signal from the farther transmitter 200 microseconds later than from the nearer one. If signal speed is 300 km/microsecond, find the equation of the hyperbola on which the ship lies.

Solution

Step 1: Calculate distance difference: Time difference = 200 microseconds Signal speed = 300 km/microsecond Distance difference = 300 × 200 = 60,000 km (This is unrealistic; rescale for practical units) Let's assume the distance difference = 60 km. For a hyperbola: |d₁ - d₂| = 2a 60 = 2a a = 30 km Step 2: Identify foci: Foci: (-100, 0) and (100, 0) c = 100 km Step 3: Calculate b: c² = a² + b² (100)² = (30)² + b² 10000 = 900 + b² b² = 9100 b = √9100 ≈ 95.39 km Step 4: Write hyperbola equation: Center: (0, 0) [midpoint of foci] Transverse axis: horizontal (foci on x-axis) (x)²/a² - (y)²/b² = 1 (x)²/900 - (y)²/9100 = 1 or equivalently: x²/900 - y²/9100 = 1 Step 5: Eccentricity: e = c/a = 100/30 = 10/3 ≈ 3.33 Answer: x²/900 - y²/9100 = 1 The ship is located somewhere on this hyperbola.

Key Points

  • Standard form (horizontal): (x - h)²/a² - (y - k)²/b² = 1
  • Standard form (vertical): (y - k)²/a² - (x - h)²/b² = 1
  • Center: (h, k)
  • Vertices: distance a from center along transverse axis
  • Foci: distance c from center, where c² = a² + b²
  • Eccentricity: e = c/a > 1 (always > 1 for hyperbola)
  • Defining property: |distance to focus₁ − distance to focus₂| = 2a
  • Asymptotes: y - k = ±(b/a)(x - h) for horizontal hyperbola
  • Asymptotes: y - k = ±(a/b)(x - h) for vertical hyperbola
  • Two separate branches on opposite sides of center
  • Used in navigation systems (LORAN) and structure design

The four conic sections are classified by their eccentricity (e), a parameter that measures how much the curve deviates from a circle. Understanding which type of conic you're dealing with is essential for solving complex engineering problems. **Eccentricity Classification:** 1. **Circle: e = 0** - Equation: (x - h)² + (y - k)² = r² - All points are equidistant from a single center - Used for roundabouts, circular curves, areas of influence 2. **Ellipse: 0 < e < 1** - Equation: (x - h)²/a² + (y - k)²/b² = 1 (a ≠ b) - Two foci; sum of distances to foci is constant - c² = a² - b² - Used for transition curves, architectural elements 3. **Parabola: e = 1** - Equation: (y - k)² = 4a(x - h) or (x - h)² = 4a(y - k) - One focus and directrix; distances are equal - Most important for highway vertical alignment (NSCP 2015) - Used for arches, suspensions, drainage paths 4. **Hyperbola: e > 1** - Equation: (x - h)²/a² - (y - k)²/b² = 1 (or with y first) - Two foci; difference of distances to foci is constant - c² = a² + b² - Used for navigation systems, structural analysis **Quick Identification From Equation:** - If x² and y² have the same sign and same coefficient → **Circle** - If x² and y² have the same sign but different coefficients → **Ellipse** - If x² and y² have opposite signs (one + and one −) → **Hyperbola** - If only one squared term (x² or y², not both) → **Parabola** **Relationship Between Ellipse and Hyperbola:** - Ellipse: c² = a² - b² (subtract) - Hyperbola: c² = a² + b² (add) - This is the key difference in calculating focal distance

Heading

8. Summary of Conic Sections and Eccentricity Classification

Examples

Problem

Classify each conic and find its eccentricity: (a) x² + y² - 4x + 6y - 3 = 0 (b) x²/25 + y²/9 = 1 (c) y² = 12x (d) x²/16 - y²/20 = 1

Solution

(a) x² + y² - 4x + 6y - 3 = 0 Step 1: Identify coefficients of squared terms Both x² and y² present with same coefficient (1) Classification: CIRCLE Step 2: Complete the square to find center and radius (x² - 4x) + (y² + 6y) = 3 (x - 2)² - 4 + (y + 3)² - 9 = 3 (x - 2)² + (y + 3)² = 16 Center: (2, -3), Radius: 4 Eccentricity: e = 0 (b) x²/25 + y²/9 = 1 Step 1: Identify form Positive signs, different denominators Classification: ELLIPSE Step 2: Identify a and b a² = 25 → a = 5 (larger, semi-major axis) b² = 9 → b = 3 (smaller, semi-minor axis) Step 3: Calculate c c² = a² - b² = 25 - 9 = 16 c = 4 Eccentricity: e = c/a = 4/5 = 0.8 (c) y² = 12x Step 1: Check for squared terms Only y² present; x appears to first power Classification: PARABOLA Step 2: Convert to standard form y² = 12x (y - 0)² = 4(3)(x - 0) 4a = 12 → a = 3 Vertex: (0, 0) Focus: (3, 0) Eccentricity: e = 1 (d) x²/16 - y²/20 = 1 Step 1: Identify form Opposite signs (− between terms) Classification: HYPERBOLA Step 2: Identify a and b a² = 16 → a = 4 b² = 20 → b = √20 = 2√5 Step 3: Calculate c c² = a² + b² = 16 + 20 = 36 c = 6 Eccentricity: e = c/a = 6/4 = 1.5 Answer: (a) Circle, e = 0 (b) Ellipse, e = 0.8 (c) Parabola, e = 1 (d) Hyperbola, e = 1.5

Problem

An engineer must distinguish between three structures: (A) a circular water tank, (B) an elliptical plaza, and (C) a parabolic arch. Given three equations, classify each and state the engineering application.

Solution

Given equations (assumed): (A) (x - 50)² + (y - 50)² = 225 (B) (x - 100)²/400 + (y - 100)²/225 = 1 (C) (x - 75)² = 40(y - 120) Classification and Application: (A) (x - 50)² + (y - 50)² = 225 Type: CIRCLE Center: (50, 50) Radius: 15 m Application: Water storage tank with circular footprint (B) (x - 100)²/400 + (y - 100)²/225 = 1 Type: ELLIPSE a = 20 m, b = 15 m Center: (100, 100) Application: Elliptical plaza design (C) (x - 75)² = 40(y - 120) Type: PARABOLA Vertex: (75, 120) 4a = 40 → a = 10 m Application: Parabolic arch for coverage or structural support Answer: See classifications above—each serves distinct civil engineering purposes.

Key Points

  • Circle: e = 0, equation (x - h)² + (y - k)² = r²
  • Ellipse: 0 < e < 1, equation (x - h)²/a² + (y - k)²/b² = 1
  • Parabola: e = 1, equation (y - k)² = 4a(x - h) or (x - h)² = 4a(y - k)
  • Hyperbola: e > 1, equation (x - h)²/a² - (y - k)²/b² = 1
  • Eccentricity determines conic type: 0, between 0-1, 1, or >1
  • Ellipse focal relationship: c² = a² - b²
  • Hyperbola focal relationship: c² = a² + b²
  • Quick ID: same sign and same coeff → circle; same sign, diff coeff → ellipse; opposite signs → hyperbola; one squared term → parabola
  • Parabola most important for civil engineering (highway curves)
  • All four conics have practical applications in civil engineering
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