Porisma

Conic sections

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By the end, you can

  • Identify parabolas, ellipses and hyperbolas
  • Find centres, intercepts and asymptotes
  • Preserve restrictions when interpreting a conic as a function

about 20 minutes for the lesson · practise at your own pace.

The central idea

Conic equations describe curves that may fail the vertical-line test. In standard form, a sum of positive squared terms gives an ellipse; a difference gives a hyperbola. Completing the square identifies a shifted centre. A parabola has one squared coordinate and one linear coordinate.

A plane cuts a double cone to give four familiar conics.

Parallel to the circular base, it gives a circle. A tilted plane cutting just one nappe (half) without becoming parallel to a generator gives an ellipse.

A plane parallel to a generator gives a parabola.

A plane perpendicular to the base and cutting both nappes gives the two branches of a hyperbola; on one nappe you see one branch.

The cuts here avoid the cone’s vertex, so they are non-degenerate curves.

The source motivates cones with ice-cream and traffic cones, ellipses with planetary orbits, parabolas with a thrown ball or fountain jet, and hyperbolas with a lampshade’s light on a wall.

These are geometric models: projectile paths are approximately parabolic when drag is neglected, and the light pattern depends on the wall’s cut through the light cone.

The source also mentions Earth as an ellipsoid, the three-dimensional analogue of an ellipse.

To begin exploring the shapes, the Teacher copy suggests video prompt 1 and video prompt 2. The printed QR points to the conic visualization and applications resource. These are the source’s learning pointers; their external contents are not reproduced here. The original printed illustration credits the University of Georgia conics introduction; the figure below is independently reconstructed.

Four plane cuts of a double cone

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Four 3D schematics: a plane parallel to the base gives a circle; a tilted plane through one nappe gives an ellipse; a plane parallel to a generator gives a parabola; a vertical plane through both nappes gives two hyperbola branches. Blue marks each intersection and orange its cutting plane. The finite cones truncate open curves; circles appear foreshortened in the 3D projection.
Four 3D schematics: a plane parallel to the base gives a circle; a tilted plane through one nappe gives an ellipse; a plane parallel to a generator gives a parabola; a vertical plane through both nappes gives two hyperbola branches. Blue marks each intersection and orange its cutting plane. The finite cones truncate open curves; circles appear foreshortened in the 3D projection.
Four 3D schematics: a plane parallel to the base gives a circle; a tilted plane through one nappe gives an ellipse; a plane parallel to a generator gives a parabola; a vertical plane through both nappes gives two hyperbola branches. Blue marks each intersection and orange its cutting plane. The finite cones truncate open curves; circles appear foreshortened in the 3D projection.
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Vertical parabola sign cases

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Vertical parabolas in both sign cases: positive a opens up, negative a opens down. Basic vertex (0,0), symmetry x=0.
Vertical parabolas in both sign cases: positive a opens up, negative a opens down. Basic vertex (0,0), symmetry x=0.
Vertical parabolas in both sign cases: positive a opens up, negative a opens down. Basic vertex (0,0), symmetry x=0.
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Vertical parabola translations

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Vertical parabolas in both sign cases: positive a opens up, negative a opens down. Translated vertex (h,k), symmetry x=h. Illustration h=2,k=1.
Vertical parabolas in both sign cases: positive a opens up, negative a opens down. Translated vertex (h,k), symmetry x=h. Illustration h=2,k=1.
Vertical parabolas in both sign cases: positive a opens up, negative a opens down. Translated vertex (h,k), symmetry x=h. Illustration h=2,k=1.
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Horizontal parabola sign cases

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Horizontal parabolas in both sign cases: positive a opens right, negative a opens left. Basic vertex (0,0), symmetry y=0.
Horizontal parabolas in both sign cases: positive a opens right, negative a opens left. Basic vertex (0,0), symmetry y=0.
Horizontal parabolas in both sign cases: positive a opens right, negative a opens left. Basic vertex (0,0), symmetry y=0.
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Horizontal parabola translations

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Horizontal parabolas in both sign cases: positive a opens right, negative a opens left. Translated vertex (h,k), symmetry y=k. Illustration h=2,k=1.
Horizontal parabolas in both sign cases: positive a opens right, negative a opens left. Translated vertex (h,k), symmetry y=k. Illustration h=2,k=1.
Horizontal parabolas in both sign cases: positive a opens right, negative a opens left. Translated vertex (h,k), symmetry y=k. Illustration h=2,k=1.
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From a unit circle to an ellipse

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Three stages: unit circle, horizontal stretch by positive a replacing x with x/a, then vertical stretch by positive b replacing y with y/b. Result x²/a²+y²/b²=1. Example factors a=2,b=3; equal axis scales.
Three stages: unit circle, horizontal stretch by positive a replacing x with x/a, then vertical stretch by positive b replacing y with y/b. Result x²/a²+y²/b²=1. Example factors a=2,b=3; equal axis scales.
Three stages: unit circle, horizontal stretch by positive a replacing x with x/a, then vertical stretch by positive b replacing y with y/b. Result x²/a²+y²/b²=1. Example factors a=2,b=3; equal axis scales.
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Ellipse axes and translation

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Two panels: origin-centred ellipse with A(a,0), A′(−a,0), B(0,b), B′(0,−b); lengths 2a,2b when a>b. Translated ellipse centred at (h,k), symmetry x=h,y=k, endpoints h±a,k±b and semi-axes a,b. Numerical examples explicitly labelled.
Two panels: origin-centred ellipse with A(a,0), A′(−a,0), B(0,b), B′(0,−b); lengths 2a,2b when a>b. Translated ellipse centred at (h,k), symmetry x=h,y=k, endpoints h±a,k±b and semi-axes a,b. Numerical examples explicitly labelled.
Two panels: origin-centred ellipse with A(a,0), A′(−a,0), B(0,b), B′(0,−b); lengths 2a,2b when a>b. Translated ellipse centred at (h,k), symmetry x=h,y=k, endpoints h±a,k±b and semi-axes a,b. Numerical examples explicitly labelled.
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Horizontal hyperbola and translation

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Two panels: x²/a²−y²/b²=1 has centre(0,0), vertices (±a,0), symmetry x=0,y=0 and asymptotes y=±(b/a)x. Translation by h,k gives centre(h,k), vertices (h±a,k), symmetry x=h,y=k and asymptotes y=k±(b/a)(x−h). Source illustration assumes h>0,k>0; figures explicitly use a=2,b=1,h=4,k=2.
Two panels: x²/a²−y²/b²=1 has centre(0,0), vertices (±a,0), symmetry x=0,y=0 and asymptotes y=±(b/a)x. Translation by h,k gives centre(h,k), vertices (h±a,k), symmetry x=h,y=k and asymptotes y=k±(b/a)(x−h). Source illustration assumes h>0,k>0; figures explicitly use a=2,b=1,h=4,k=2.
Two panels: x²/a²−y²/b²=1 has centre(0,0), vertices (±a,0), symmetry x=0,y=0 and asymptotes y=±(b/a)x. Translation by h,k gives centre(h,k), vertices (h±a,k), symmetry x=h,y=k and asymptotes y=k±(b/a)(x−h). Source illustration assumes h>0,k>0; figures explicitly use a=2,b=1,h=4,k=2.
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Vertical hyperbola and translation

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Basic vertical hyperbola y²/b²−x²/a²=1 has centre(0,0), vertices (0,±b), symmetry x=0,y=0, asymptotes y=±(b/a)x. Translated form has centre(h,k), vertices (h,k±b), symmetry x=h,y=k and asymptotes y=k±(b/a)(x−h). Illustrations use a=3,b=1 and, in the translated case, positive h=3,k=2.
Basic vertical hyperbola y²/b²−x²/a²=1 has centre(0,0), vertices (0,±b), symmetry x=0,y=0, asymptotes y=±(b/a)x. Translated form has centre(h,k), vertices (h,k±b), symmetry x=h,y=k and asymptotes y=k±(b/a)(x−h). Illustrations use a=3,b=1 and, in the translated case, positive h=3,k=2.
Basic vertical hyperbola y²/b²−x²/a²=1 has centre(0,0), vertices (0,±b), symmetry x=0,y=0, asymptotes y=±(b/a)x. Translated form has centre(h,k), vertices (h,k±b), symmetry x=h,y=k and asymptotes y=k±(b/a)(x−h). Illustrations use a=3,b=1 and, in the translated case, positive h=3,k=2.
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Reciprocal model for the appendix

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The rectangular hyperbola y=1/x, equivalently xy=1, has branches in quadrants I and III and asymptotes x=0,y=0. These asymptotes are approached in limits; they contain no curve points.
The rectangular hyperbola y=1/x, equivalently xy=1, has branches in quadrants I and III and asymptotes x=0,y=0. These asymptotes are approached in limits; they contain no curve points.
The rectangular hyperbola y=1/x, equivalently xy=1, has branches in quadrants I and III and asymptotes x=0,y=0. These asymptotes are approached in limits; they contain no curve points.
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Explore

Use the worked example below, then change one input or assumption. Predict which part of the answer must change before calculating.

Related interactive example

Explore this related example, then return to the prediction task above.

Squash, slide right, slide down

Three steps turn the wide ellipse into a circle of radius 3 centred at (4, −7).

Teal: C₁: x²/36 + y²/9 = 1. No comparison overlay.

Start: C₁: x²/36 + y²/9 = 1

The original ellipse: semi-axes 6 and 3, centred at the origin.

  • Shape at the selected step
  • Centre of the current shape
Ideas to try
  1. Which step makes the ellipse round? Why does only k=12k=\frac12 work?
  2. Check the equation of the final circle against the centre (4,−7)(4,-7).
Open the full interactive problem →

Worked example

For x2/9+y2/4=1x^2/9+y^2/4=1, the centre is (0,0)(0,0), horizontal intercepts are (±3,0)(\pm3,0) and vertical intercepts (0,±2)(0,\pm2). Solving for y produces two branches y=±21−x2/9y=\pm2\sqrt{1-x^2/9} for −3≤x≤3-3\leq x\leq3.

The denominator is the square of a semi-axis length, not the length itself.

Check your understanding

Find the asymptotes of x2/9−y2/4=1x^2/9-y^2/4=1.

Mastery learning objectives

At the end of this chapter, I should be able to:

General quadratic form

Rewrite Ax2+Bx+Cy2+Dy+E=0Ax^2+Bx+Cy^2+Dy+E=0 into a standard form: parabola, ellipse, circle or hyperbola.

Parabola: y=a(x−h)2+ky=a(x-h)^2+k

Identify line of symmetry x=hx=h.

Identify vertex/turning point (h,k)(h,k).

Sketch the graph, including any necessary features required by the question.

Parabola: (y−k)2=a(x−h)(y-k)^2=a(x-h)

Identify line of symmetry y=ky=k.

Identify vertex (h,k)(h,k).

Sketch the graph, including any necessary features required by the question.

Ellipse: (x−h)2/a2+(y−k)2/b2=1(x-h)^2/a^2+(y-k)^2/b^2=1

Identify centre (h,k)(h,k).

Identify length of its major axis.

Identify length of its minor axis.

Identify vertices (h−a,k)(h-a,k), (h+a,k)(h+a,k), (h,k−b)(h,k-b) and (h,k+b)(h,k+b).

Identify lines of symmetry x=hx=h and y=ky=k.

Sketch the graph, including any necessary features required by the question.

Circle: (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2

Identify centre (h,k)(h,k).

Identify radius rr.

Identify lines of symmetry.

Sketch the graph, including any necessary features required by the question.

Horizontal hyperbola: (x−h)2/a2−(y−k)2/b2=1(x-h)^2/a^2-(y-k)^2/b^2=1

Identify centre (h,k)(h,k).

Identify vertices (h−a,k)(h-a,k) and (h+a,k)(h+a,k).

Identify lines of symmetry x=hx=h and y=ky=k.

Identify oblique asymptotes y=±(b/a)(x−h)+ky=\pm(b/a)(x-h)+k.

Sketch the graph, including any necessary features required by the question.

Vertical hyperbola: (y−k)2/b2−(x−h)2/a2=1(y-k)^2/b^2-(x-h)^2/a^2=1

Identify centre (h,k)(h,k).

Identify vertices (h,k−b)(h,k-b) and (h,k+b)(h,k+b).

Identify lines of symmetry x=hx=h and y=ky=k.

Identify oblique asymptotes y=±(b/a)(x−h)+ky=\pm(b/a)(x-h)+k.

Sketch the graph, including any necessary features required by the question.

Choose a practice question

The printed skill map (printed page 138) is retained below. A dash represents a blank source cell. Some Discussion subpart labels do not match the questions printed on the following page; the study links below use the actual questions.

Skill/conceptLecture examples, as printedDiscussion questions, as printedReview problems, as printed
Sketch parabola11(a)—
Sketch ellipse/circle41(b)(c), 2, 3 (1st part)5
Sketch hyperbola5, 71(d)(e)1, 6(a)
Sketch a conic on an existing curve to aid in solving the question—3 (last part)2(iii), 3, 4(i)
Using gradient concept to aid in solving the question—4(a)(iii)—
Prior topic: Graphing Techniques—4(a)(i)(ii)(b)2(i)(ii), 4(ii), 6(b)
Prior topic: Transformations——2(iv)

For the actual Discussion 1 equations, (a) and (b) are parabolas, (c) and (d) are ellipses, and (e) and (f) are hyperbolas. Discussion 3’s first graph is a rational curve, followed by an ellipse comparison. These qualifications preserve the printed map without directing you to the wrong conic.

Practice questions

Practice section 1 · Questions 1–6

Two parabola sketches: algebra and calculator branches · Read a parabola’s vertex and choose an equation · +4 more

Practice section 2 · Questions 7–11

Exact intercepts of a translated vertical hyperbola · Six conics: complete the square and identify graph features · +3 more

More practice

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