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Lesson 5 of 5

Conic Sections · Lesson 5 of 5

Chapter Summary and Practice

“Connect every conic, choose equations from geometry, and solve mixed modelling and reasoning problems.”

Learning Objectives

• Connect plane sections and distance definitions with the equations of all four conics. • Compare axes, vertices, foci, eccentricity and latus rectum without mixing parameter conventions. • Choose a coordinate origin and orientation that simplify an application. • Model reflector widths, beam deflection, arches, suspension cables and moving rods with consistent units. • Solve mixed geometric problems and check signs, branches, domains and physical restrictions.

One chapter, four distance rules

The four conics are related as plane sections of a double cone, but their distance conditions explain why their equations differ. Revising the chapter means reconnecting these conditions, not merely collecting formulas. When a problem describes a curved path, first ask what remains fixed: one distance, an equality, a sum or an absolute difference.

Keep the coordinate conventions visible. A circle may have any centre (h, k). The parabola forms here have their vertex at the origin, while the ellipse and hyperbola forms have their centre at the origin and their principal axes on the coordinate axes. In an application we are free to choose such an origin when the geometry permits; choosing it is part of the reasoning.

ConicDefining geometryRepresentative equationEssential parameters
CircleDistance from centre is r(x − h)² + (y − k)² = r²Centre (h, k); r > 0
ParabolaDistance to focus equals perpendicular distance to directrixy² = 4ax, or reflected/exchanged formVertex (0, 0); a > 0; focus (a, 0) in this form
EllipseSum of focal distances is 2ax²/a² + y²/b² = 1, a > b > 0a² = b² + c²; e = c/a < 1
HyperbolaAbsolute difference of focal distances is 2ax²/a² − y²/b² = 1c² = a² + b²; e = c/a > 1

For the cone picture, a plane avoiding the vertex gives a circle at β = 90°, an ellipse for α < β < 90°, a parabola for β = α, and a hyperbola for 0 ≤ β < α. Through the vertex the corresponding cases collapse to a point, a generator line, or two intersecting lines. The ordinary distance-defined curves require nondegenerate parameter choices.

A whole-chapter method guide

Different data suggest different first steps. Centre and radius lead directly to a circle equation. A focus and a directrix suggest a parabola. Two foci do not by themselves identify a conic: the sum or difference condition decides between ellipse and hyperbola. Before calculating, label what each given length represents.

Given informationFirst moveCheck before finishing
Circle equation in expanded formNormalise quadratic coefficients and complete both squaresRight side is positive; radius is its square root
Circle through points with a constrained centreSet centre (h, k); equate squared radii and use the centre conditionEvery given point and centre constraint hold
Standard parabola equationMatch the linear coefficient with ±4aa positive; squared variable identifies the other axis
Ellipse equationNormalise to 1; larger denominator is a²c² = a² − b²; 0 ≤ e < 1
Hyperbola equationNormalise to 1; positive term has denominator a²c² = a² + b²; e > 1; both branches included
A real shape with widths, depths or heightsChoose vertex or centre as origin and use a known pointUnits consistent; distinguish full width from half-width
A moving point constrained by a rigid rodExpress x and y using the rod angle or endpoint geometryState whether all quadrants or only a restricted arc are allowed
FeatureParabola y² = 4axEllipse x²/a² + y²/b² = 1Hyperbola x²/a² − y²/b² = 1
Main axisx-axis, y = 0Major axis: x-axisTransverse axis: x-axis
Vertex or vertices(0, 0)(±a, 0)(±a, 0)
Focus or foci(a, 0)(±c, 0), c² = a² − b²(±c, 0), c² = a² + b²
Other defining objectDirectrix x = −aConstant sum 2aConstant absolute difference 2a
Full latus rectum4a2b²/a2b²/a
Coordinates allowedx ≥ 0, y unrestricted|x| ≤ a, |y| ≤ b|x| ≥ a, y unrestricted
SymmetryAbout x-axisAbout both axesAbout both axes

For vertical orientation exchange x and y in these equations and coordinate descriptions. A leftward or downward parabola needs a negative linear coefficient. The ellipse’s axes come from the larger denominator, while the hyperbola’s transverse direction comes from the positive term. These selection rules must remain separate.

Mixed equation checks

Before modelling a physical situation, practise recognising the curve behind an equation. A quick substitution or parameter check is often enough to catch an error. It is especially useful when a correct-looking formula has been applied with the wrong axis or with a full length used as a half-length.

Example — Recover a circle and classify a point

Problem
Find the centre and radius of x² + y² − 4x + 6y − 12 = 0, then classify (5, −3).

  1. 1.Rearrange: (x² − 4x) + (y² + 6y) = 12. Add 4 and 9 to both sides.
  2. 2.(x − 2)² + (y + 3)² = 25. The centre is (2, −3) and radius 5.
  3. 3.For (5, −3), the squared distance to the centre is 3² + 0² = 9 < 25, so the point is inside.
Example — Same denominators, different distance rules

Problem
Compare x²/25 + y²/9 = 1 with x²/25 − y²/9 = 1.

  1. 1.For the ellipse a = 5, b = 3, c = √(25 − 9) = 4. Its foci are (±4, 0), fixed sum 10, and eccentricity 4/5.
  2. 2.For the hyperbola a = 5, b = 3, c = √(25 + 9) = √34. Its foci are (±√34, 0), fixed absolute difference 10, and eccentricity √34/5 > 1.
  3. 3.The ellipse is bounded with |x| ≤ 5. The hyperbola has two branches with |x| ≥ 5. Both have latus rectum length 18/5, but that shared formula does not make their geometry the same.
Example — Reconstruct a vertical ellipse

Problem
An origin-centred ellipse has full major-axis length 10, vertical foci and eccentricity 3/5. Find its equation.

  1. 1.Full major length 10 means a = 5. The eccentricity gives c = ae = 3.
  2. 2.Use b² = 25 − 9 = 16.
  3. 3.The major axis is vertical, so x²/16 + y²/25 = 1. Foci (0, ±3) and vertices (0, ±5) verify the data.

Parabolic mirrors and reflectors

A reflector problem gives a parabolic cross-section, not a circle. Put the vertex at the origin and take the opening direction as the positive x-axis. A depth measured along that axis gives x; a width across it gives twice the positive y-coordinate. The focal distance fixes a, allowing width and depth to determine each other.

FocusVertexABFull width ABDepth measured along x
Depth and full width of a parabolic section— At the rim, the top and bottom ordinates are opposites; the full width is 2|y|.
Example — Width of a deep parabolic mirror

Problem
A parabolic mirror has focus 5 cm from its vertex and depth 45 cm. Find its rim width.

  1. 1.Take the vertex as origin and the axis along positive x. Since a = 5 cm, y² = 4ax = 20x.
  2. 2.At the rim x = 45 cm, so y² = 20 × 45 = 900 and y = ±30 cm.
  3. 3.The upper and lower rim points are (45, 30) and (45, −30). Their separation is 60 cm.
  4. 4.Using just the positive ordinate would give half the width. Depth and focal distance were both expressed in centimetres.
Example — Locate a reflector’s focus

Problem
A parabolic reflector has diameter 20 cm and depth 5 cm. Find its focus relative to the vertex.

  1. 1.Use y² = 4ax. The rim has half-diameter |y| = 10 cm and axial depth x = 5 cm.
  2. 2.Substitute: 100 = 4a × 5 = 20a, so a = 5 cm.
  3. 3.The focus is (5, 0) in these coordinates: 5 cm along the opening axis from the vertex. The diameter must be halved before substituting y.

Height is not always deflection

In a deflected beam, choose the lowest point as the parabola’s vertex. Its y-coordinate then measures height above the lowest point. Deflection is usually measured downward from the original horizontal support level. These two quantities add to the total central deflection; confusing them changes the answer.

Support level: 3 cm above vertexO1 cm deflection2 cm above vertex
Beam height versus downward deflection— The vertical scale is exaggerated to make the small deflection visible; the lowest point is the origin.
Example — A beam deflected at its centre

Problem
A beam is supported at endpoints 12 m apart. It sags 3 cm at its centre and is modelled as a parabola. How far from the centre is the deflection 1 cm?

  1. 1.Put the vertex at the lowest point and the y-axis vertically upward. The supports have coordinates (±6, 0.03) in metres.
  2. 2.Use x² = 4ay. At a support, 36 = 4a × 0.03 = 0.12a, so a = 300 m.
  3. 3.A deflection of 1 cm below the support level means height y = 3 cm − 1 cm = 2 cm = 0.02 m above the vertex.
  4. 4.Substitute: x² = 4 × 300 × 0.02 = 24. Thus x = ±2√6 m.
  5. 5.The requested distance from the centre is |x| = 2√6 m, approximately 4.90 m, at two symmetric positions. The large focal parameter is compatible with a broad, shallow parabola.

Arches and suspended cables

For a downward-opening arch, choose the top as the vertex and use y upward. Points below it then have negative y-coordinates. For a suspended cable, choose its lowest point as the vertex; the roadway may lie below this origin, so wire length is an offset plus the cable’s height. These coordinate choices simplify equations while keeping the physical meaning visible.

Example — Width of a parabolic arch

Problem
An arch has vertical axis, height 10 m and base width 5 m. How wide is it 2 m below its vertex?

  1. 1.Choose the top as (0, 0), with y upward. The equation is x² = −4ay. Base endpoints are (±2.5, −10).
  2. 2.Substitution gives 6.25 = 40a, so a = 5/32 m and 4a = 5/8 m.
  3. 3.At y = −2, x² = −(5/8)(−2) = 5/4. Hence x = ±√5/2.
  4. 4.Full width is √5 m, approximately 2.24 m. This is smaller than the 5 m base width, as a section closer to the top should be.
Example — A supporting wire away from the bridge centre

Problem
A parabolic suspension cable spans a horizontal 100 m roadway. The shortest central supporting wire is 6 m, and the longest endpoint wires are 30 m. Find the wire length 18 m from the middle.

  1. 1.Place the cable’s lowest point at (0, 0). The roadway is y = −6, so a wire at x has length y + 6.
  2. 2.Endpoint coordinates of the cable are (±50, 24), since 30 − 6 = 24 m above the lowest cable point.
  3. 3.Use y = Kx². Then 24 = K × 2500, giving K = 24/2500.
  4. 4.At x = 18, y = (24/2500) × 324 = 3.1104 m.
  5. 5.The wire length is 3.1104 + 6 = 9.1104 m. The same length occurs 18 m on either side; it lies between 6 and 30 m as expected.
Example — Height of a semi-elliptical arch

Problem
An arch is the upper half of an ellipse, 8 m wide and 2 m high at the centre. Find its height 1.5 m from one end.

  1. 1.Choose the midpoint of the base as the origin. The horizontal half-width is a = 4 m and vertical half-height is b = 2 m.
  2. 2.The ellipse is x²/16 + y²/4 = 1, restricted to y ≥ 0 for the arch.
  3. 3.A point 1.5 m from the left end x = −4 has x = −2.5. Using the right end gives x = 2.5 and the same height by symmetry.
  4. 4.Then y² = 4(1 − 6.25/16) = 39/16. Since the arch is above the base, y = √39/4 m, approximately 1.56 m.
  5. 5.Reject the negative square root for the physical arch, even though it lies on the full algebraic ellipse.

A moving rod generates an ellipse

A rigid rod can link a moving point to both coordinate axes. Its length supplies a fixed geometric relationship. Express the point’s coordinates using the rod’s angle, then eliminate that angle with cos²θ + sin²θ = 1. The result is an ellipse equation, even though no foci were mentioned in the problem.

B on y-axisA on x-axisP(x, y)AP = 6PB = 9Rod length AB = 15; x/9 = cos θ, y/6 = sin θ
Coordinates of a point on a sliding rod— The horizontal coordinate uses PB and the vertical coordinate uses AP; these are easy to interchange accidentally.
Example — A point six centimetres from one end

Problem
A rod AB has length 15 cm, with A on the x-axis and B on the y-axis. A point P is on the rod with AP = 6 cm. Find its locus.

  1. 1.Since AB = 15 and AP = 6, PB = 9 cm. Let θ be the acute angle the first-quadrant rod makes with the horizontal.
  2. 2.The horizontal projection from B to P gives x = 9 cos θ. The vertical projection from A to P gives y = 6 sin θ.
  3. 3.Thus x/9 = cos θ and y/6 = sin θ. Squaring and adding yields x²/81 + y²/36 = 1.
  4. 4.This is an ellipse with horizontal semi-major axis 9 and semi-minor axis 6. If both endpoints are constrained to the positive coordinate rays, only its first-quadrant arc is traced.
  5. 5.If the endpoints may move on the full axes, sign changes allow the full ellipse. State the allowed motion rather than assuming that a physical arrangement traces every equation point.
Example — A different division point on a rod

Problem
A rod of length 12 cm touches the coordinate axes. P is 3 cm from the end on the x-axis. Determine its locus.

  1. 1.Let A be the x-axis end and B the y-axis end. AP = 3 and PB = 9.
  2. 2.The same projection reasoning gives x = 9 cos θ and y = 3 sin θ.
  3. 3.Therefore x²/81 + y²/9 = 1. The longer semi-axis is the remaining distance PB, not the total rod length.
  4. 4.With positive-ray motion, restrict to the first-quadrant arc; with unrestricted axis motion, the locus extends over the whole ellipse.

Distance loci and triangle geometry

The final problems combine the defining distances with familiar geometry. A constant sum from two posts immediately suggests an ellipse. A triangle involving a parabola’s latus rectum requires its endpoints before area can be calculated. Symmetry can also reduce a more difficult inscribed-triangle problem to one variable.

Example — A runner’s constant-sum path

Problem
A runner keeps the sum of distances from two flagposts equal to 10 m. The posts are 8 m apart. Find the path’s equation.

  1. 1.Take the midpoint between the posts as the origin and their joining line as the x-axis. The posts are the foci (±4, 0), so c = 4.
  2. 2.The distance sum is 2a = 10, giving a = 5. Then b² = a² − c² = 25 − 16 = 9.
  3. 3.The path is x²/25 + y²/9 = 1 in these coordinates.
  4. 4.At (0, 3), each focal distance is √(16 + 9) = 5, checking the total 10. The runner’s position moves along the ellipse; the posts themselves are fixed.
Example — Triangle formed by a parabolic focal chord

Problem
Find the area of the triangle joining the vertex of x² = 12y to the endpoints of its latus rectum.

  1. 1.Compare x² = 4ay: a = 3. The focus is (0, 3), and the latus rectum lies on y = 3.
  2. 2.At y = 3, x² = 36, so endpoints are (−6, 3) and (6, 3). The base length is 12.
  3. 3.The vertex is (0, 0), whose perpendicular distance to the base line y = 3 is 3.
  4. 4.Area = (1/2) × 12 × 3 = 18 square units. A side from the vertex to an endpoint is not the triangle’s height.
Example — An equilateral triangle inscribed in a parabola

Problem
An equilateral triangle is inscribed in y² = 4ax, a > 0, with one vertex at the parabola’s vertex. Find its side length.

  1. 1.Call the vertex O = (0, 0) and the other points P and Q. Since OP = OQ, their squared distances from O are equal.
  2. 2.On the parabola, OP² = x² + y² = x² + 4ax. For x ≥ 0 this expression is strictly increasing in x, so equal distances force P and Q to have the same x-coordinate.
  3. 3.Their y-coordinates must therefore be opposites: write P = (u, t) and Q = (u, −t), with u, t > 0. The base PQ has length 2t and OP has length √(u² + t²).
  4. 4.Equilateral sides require √(u² + t²) = 2t, so u² = 3t² and u = √3t.
  5. 5.Use the parabola condition t² = 4au = 4a√3t. Divide by t > 0 to get t = 4√3a, and hence u = 12a.
  6. 6.The side length is 2t = 8√3a. Check OP² = (12a)² + (4√3a)² = 192a², whose positive square root is also 8√3a.

What the derivations teach us

All four equations come from distances, but squaring can hide signs. The circle and parabola converses compare nonnegative distances directly. For the ellipse, |x| ≤ a keeps a ± cx/a positive. For the hyperbola, the signs differ on the two branches and must be resolved separately. These checks explain why an algebraic equation truly represents the intended geometric locus.

Symmetry is also an algebraic idea: replacing a coordinate by its negative leaves a squared term unchanged. Yet the physical model may use only part of that symmetric curve. A semi-elliptical arch uses the upper half; a rod on positive rays uses a quarter; a mirror uses a finite section of an otherwise unbounded parabola. Always distinguish the mathematical curve from the permitted physical region.

Common errors across the chapter

Avoid using a full width as a coordinate, mixing centimetres with metres, confusing deflection with height above a vertex, treating an ellipse’s foci as vertices, using the larger-denominator rule for hyperbolas, or dropping absolute values too early. For a model, also ask whether every algebraic branch or sign is physically allowed.

How geometry and algebra came together

Conics were studied geometrically long before coordinates became a standard method. Greek work, particularly that of Apollonius and Archimedes, explored their properties as curves. Earlier Indian and Greek geometric traditions developed reasoning about figures without routinely turning every figure into an algebraic equation. Coordinates later made it possible to move between a geometric condition and an equation systematically.

A historical connection

Descartes’s La Géométrie appeared in 1637, while Fermat developed related methods independently; his work on loci was published after his death in 1679. Later mathematicians helped establish the coordinate language and formulas used today. The key shift was to represent a moving point by numbers and its geometric restrictions by equations.

The historical development continued through Barrow, Newton and Leibniz, and later writers including L’Hospital, Clairaut, Cramer, Monge and Lacroix. It involved alternative coordinate choices, distance and line formulas, systematic use of two axes, and ways to describe intersections of loci, including work by Lamé. These names provide context; the enduring idea for this chapter is that a curve can be studied both as a figure and as an equation.

Check your understanding

The questions below revisit all four conics and the modelling choices that connect them. Work from the defining relationship before choosing an equation, and check whether a length is a full span, a half-span or an offset from the chosen origin.

Quiz

Quick check

Which equation is a circle with centre (2, −1) and radius 3?

Quick check

Which distance condition describes a parabola?

Quick check

A horizontal ellipse has distance sum 10 and focal separation 8. Which equation fits?

Quick check

A hyperbola has a = 4 and b = 3. What is its eccentricity?

Quick check

At a parabolic mirror’s rim, the two ordinates are ±12 cm. What is its full width?

Quick check

A beam’s central deflection is 3 cm. At a point deflected 1 cm below support level, what is its height above the lowest point?

Quick check

A rod is 12 cm long, with P 3 cm from its x-axis end. Which locus equation follows?

Quick check

What is the area of the triangle joining the vertex of x² = 12y to its latus rectum endpoints?

Quick check

An equilateral triangle on y² = 4ax includes the parabola’s vertex. What is its side length?

Quick check

A physical arch is the upper half of an ellipse. Which restriction must be added to its full equation?

Practice Problems

Practice Problems
  1. Create a comparison of the four conics from memory: distance condition, equation, axis choice and one misconception for each.
  2. Find the centre and radius of x² + y² + 6x − 8y = 0, and classify (0, 0) and (−3, 4).
  3. For y² = −20x, find the focus, directrix, latus rectum endpoints and allowed x-region.
  4. Find all main features of x²/16 + y²/25 = 1 and compare them with y²/25 − x²/16 = 1.
  5. A parabolic reflector is 30 cm wide and 9 cm deep. Find the focus relative to its vertex, showing the half-width calculation.
  6. For the 12 m beam model with central deflection 3 cm, find the positions where the downward deflection is 2 cm. Distinguish deflection from y.
  7. Using the 100 m bridge model above, find the supporting-wire length 25 m from the middle and explain why the central wire offset must be added.
  8. A semi-elliptical arch is 12 m wide and 4 m high. Find its height 2 m from an end and state the relevant sign restriction.
  9. A rod 10 cm long moves with ends on the axes. Find the locus of a point 4 cm from its x-axis end; describe the restriction for positive-ray motion.
  10. Two foci are 12 units apart and the distance sum is 14. Choose horizontal coordinates and find the ellipse equation.
  11. Find the area of the vertex–latus rectum triangle for y² = 20x, using a base and a perpendicular height.
  12. Re-derive the side length of the equilateral triangle inscribed in y² = 4ax and explain why the two nonvertex points have opposite y-coordinates.

Key Takeaways

Key Takeaways

• The four conics are tied together by cone sections and by four distinct distance rules. • Choose the equation from the geometric condition, then identify parameters and axis direction. • Ellipse and hyperbola use different focal relationships and different rules for selecting the main axis. • Distance derivations require converses and sign checks; a squared equation alone can conceal assumptions. • In models, choose coordinates deliberately, use consistent units and distinguish full widths, half-widths and vertical offsets. • A physical setting may trace only part of the algebraic curve; verify every permitted sign and branch.

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Hyperbolas: Two Foci and a Constant Difference

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