Conic Sections

Ryan Hirst

The main body of this book --under construction-- is a complete solution to the general problem: [b]Proposition[/b] [i]To draw a conic section of which five elements − points and tangents − are known.[/i] The chapters shold cover all subproblems: 1. Drawing solution 2. Projection and involution 3. Ambiguous cases -- coincident points 4. Stable coefficients from the linear equations (matrices) 5. Manipulating the curve by matrix transform. 6. Determining the properties (focal length, major axis direction...) 7. Descriptive coefficients 8. Conversion to parametric equations (orienting the curves). I intend to include among my GGB books all the auxiliary theorems, down to Euclid, for that is how I solved the problem. Your help is appreciated. I have so far left a great may steps undocumented. Let me know where there are missing theorems, and where additional documentation will help you. This is problem #64 in Heinrich Dorrie's [i]100 Great Problems of Elementary Mathematics.[/i] The rest are assorted worksheets. Most are sub-problems to larger projects, but I try to make my worksheets self-contained. Le me know if I can make these more useful to you.

Table of Contents

  1. Five Elements I - Drawing Solution
    1. 5 Tangents -- Points of Tangency
    2. Brianchon's Hexagram Theorem
    3. 5 Tangents -- Insert a Tangent Line
    4. A Suitable Pentagram
    5. Conic, 5 Tangents - Drawing Solution
  2. Five Elements II -- Analytic
    1. Conic Section: Properties
    2. Ambiguity I - Collinear Points
    3. Conics - Translation
  3. Five Elements III - The Cases
    1. Conic Section: 3 Points, 2 Tangents
    2. Conic Section: 4 points, 1 tangent
    3. Conic Section: 5 tangents
    4. Conic Section: 5 points
  4. The Parabola
    1. Vectors: Parabola from Three Points (Focus and Directrix)
    2. Vector Parabola 3D
    3. 56. Area of a Parabola - Archimedes
    4. Bézier Parabola
    5. Parabola from Four Tangents, 2
    6. The Parabola: Focal angle of intersecting tangents
    7. Parabola from Four Tangents
    8. Parabola from Vertex and Two Points on the Arc
    9. Area Under a Parabola
  5. The Ellipse
    1. Ellipse From String and Tacks
    2. Ellipse by parameter, scale, rotation
    3. Ellipse: Foci from parametric equation
    4. Ellipse: Path Parameter, conjugate radius
  6. Rotation
    1. Ellipse: Oriented Rotation 1
    2. Ellipse: Oriented Rotation 2
    3. Conic Rotation 3: half-axes
  7. Classic Problems
    1. The Locus of Points Equidistant from Two Circles
    2. 47. Trammel! Van Schooten's Locus Problem
    3. 48. Cardan's Spur Wheel
    4. Cycloids and the Rolling Circle
    5. Archimedes' Formula for π (pi)

1.

Five Elements I - Drawing Solution

From five given tangents, a conic section can be drawn.

  • 1. 5 Tangents -- Points of Tangency
  • 2. Brianchon's Hexagram Theorem
  • 3. 5 Tangents -- Insert a Tangent Line
  • 4. A Suitable Pentagram
  • 5. Conic, 5 Tangents - Drawing Solution

1.1.

5 Tangents -- Points of Tangency

From five tangents, determine the points of tangency on each tangent line.
– Ryan Hirst

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2.

Five Elements II -- Analytic

  • 1. Conic Section: Properties
  • 2. Ambiguity I - Collinear Points
  • 3. Conics - Translation

2.1.

Conic Section: Properties

Finding a more descriptive form for the equation, and giving the curve properties.
– Ryan Hirst

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3.

Five Elements III - The Cases

  • 1. Conic Section: 3 Points, 2 Tangents
  • 2. Conic Section: 4 points, 1 tangent
  • 3. Conic Section: 5 tangents
  • 4. Conic Section: 5 points

3.1.

Conic Section: 3 Points, 2 Tangents

[b]Case:[/b] [i]Three points and two tangents are given.[/i]

NOTES: The points of tangency D, E are found using Involution (projection). I then know [i]five points,[/i] a case I can solve ([url]http://www.geogebratube.org/material/show/id/213219[/url]). When the solution exists, there are four choices. Drag D to any of the red diamonds. ________ Construction of the orange diamonds (double points of an involution): [url]http://www.geogebratube.org/material/show/id/243899[/url]
– Ryan Hirst

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4.

The Parabola

  • 1. Vectors: Parabola from Three Points (Focus and Directrix)
  • 2. Vector Parabola 3D
  • 3. 56. Area of a Parabola - Archimedes
  • 4. Bézier Parabola
  • 5. Parabola from Four Tangents, 2
  • 6. The Parabola: Focal angle of intersecting tangents
  • 7. Parabola from Four Tangents
  • 8. Parabola from Vertex and Two Points on the Arc
  • 9. Area Under a Parabola

4.1.

Vectors: Parabola from Three Points (Focus and Directrix)

Classical Parabola construction from three points F, A, B. The parabola is all points equidistant from F and the line through A and B.

More: The same equation in 3-dimensional space: [url]http://www.geogebratube.org/student/m32476[/url] The second order Bezier Curve: [url]http://www.geogebratube.org/material/show/id/37814[/url] ToDo: Show the parabola defined as a conic section. A cone touches F and the line D = AB. Move the cone up and down, through the plane, holding F and D fixed. Let F and D draw on the cone. Probably easier in Blender.
– Ryan Hirst

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5.

The Ellipse

  • 1. Ellipse From String and Tacks
  • 2. Ellipse by parameter, scale, rotation
  • 3. Ellipse: Foci from parametric equation
  • 4. Ellipse: Path Parameter, conjugate radius

5.1.

Ellipse From String and Tacks

Question: [i]Given the ellipse equation, x²/a² + y²/b²=1 How can I show that any two rays drawn from the foci and meeting at a point on the arc add up to the same length?[/i] I ask the question this way: [i]Given two tacks (foci) and a string (constant length), pull the string tight and draw. Prove that the resulting figure has the algebraic equation x²/a² + y²/b²=1.[/i] ______________________ Accompanying text: [url]http://mathosaurus.blogspot.com/2013/06/constructing-ellipse.html[/url] The equation at the top of the applet is an intermediate step.

____________ These are my self-study materials. Let me know how I can make them more useful to you.
– Ryan Hirst

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6.

Rotation

  • 1. Ellipse: Oriented Rotation 1
  • 2. Ellipse: Oriented Rotation 2
  • 3. Conic Rotation 3: half-axes

6.1.

Ellipse: Oriented Rotation 1

[b]Proposition:[/b] Given an ellipse rotated by ∡θ, determine the relationships among the values a, b, θ, and the equations i) Ax² −2Bxy + Cy² = F ii) x²/a² + y²/b² =1 iii) parametric equation r(t) = a u + b v

Notes: *Parameter F is not free. It is determined by A, B, C: F = AC - B² I used a special case of the standard form to isolate the relationship among axes, angle and coefficients. TO DO: connect it with natural solution of the differential equation dy/dx = (a1x−a2y) /(a2x + b2y), where F is free (the constant of integration) but the coefficients are determined by the angle. * -π/2 < arctan() < π/2. But θ may go outside this range. To resolve the full range 0≤ θ ≤ 2π using the standard form, I need additional information. Onward-- ________________ Ellipse Rotation, (1 of 3): [b]→1: Converting between standard form and parametric equations.[/b] 2. Resolve θ, 0 ≤ θ≤ 2π using atan(); continuity troubles: [url]http://www.geogebratube.org/material/show/id/45026[/url] 3. Determine the half-axis lengths and orientation, disambiguate tan(θ) for limit cases: [url]http://www.geogebratube.org/material/show/id/45924[/url]
– Ryan Hirst

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7.

Classic Problems

  • 1. The Locus of Points Equidistant from Two Circles
  • 2. 47. Trammel! Van Schooten's Locus Problem
  • 3. 48. Cardan's Spur Wheel
  • 4. Cycloids and the Rolling Circle
  • 5. Archimedes' Formula for π (pi)

7.1.

The Locus of Points Equidistant from Two Circles

The locus of points equidistant from two circles.

There are two solutions: a single branch of a hyperbola (orange), and an ellipse (blue). Case I: If one circle completely encloses the other, there is one solution: an ellipse. Case II: If the two circles are entirely outside one another, there is one solution: the branch of the hyperbola closer to the midpoint of the smaller circle. Case III: If the two circles intersect, both solutions occur simultaneously. Please let me know if you would like more information about the construction; some of the details are not obvious, even from the full worksheet.
– Ryan Hirst

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