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Contents · 0% complete
  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Conic sections · Subtopic 2 of 2

The Ellipse and Hyperbola

x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is an ellipse with b2=a2(1−e2)b^2=a^2(1-e^2); x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 is a hyperbola with b2=a2(e2−1)b^2=a^2(e^2-1). Both have foci (±ae,0)(\pm ae,0) and directrices x=±aex=\pm\frac ae; distance to a focus is ee times distance to the matching directrix.

Edexcel YFM01 WFM03Edexcel 9FM0 Option≈ 50 minRevision 1 · updated 2026-10-27
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Section 1 of 14

Learning objectives

  • Find eccentricity, foci and directrices.
  • Use parametric forms (acos⁡θ,bsin⁡θ)(a\cos\theta,b\sin\theta) and (asec⁡θ,btan⁡θ)(a\sec\theta,b\tan\theta).
  • Find tangents and normals; know the hyperbola's asymptotes.
Pause and think

Pin two ends of a loop of string to a board and pull a pencil round it. What shape do you draw?

Section 2 of 14

Core idea

Start hereA loop of string and two pins

Push two pins into a board, loop string round them and pull it tight with a pencil. As you draw, the two distances to the pins always add to the length of string. The curve is an ellipse, and the same idea with a difference of distances gives a hyperbola.

The big idea

Every conic has distance to focus =e×=e\times distance to directrix. For e<1e<1 you get the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 with b2=a2(1−e2)b^2=a^2(1-e^2); for e>1e>1 the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 with b2=a2(e2−1)b^2=a^2(e^2-1).

1. Finding the foci

Ellipse from the focus–directrix rule
  1. (x−ae)2+y2=e2(x−ae)2(x-ae)^2+y^2=e^2\left(x-\tfrac ae\right)^2
    PS2=e2PM2PS^2=e^2PM^2 with focus (ae,0)(ae,0) and directrix x=aex=\frac ae.
  2. x2−2aex+a2e2+y2=e2x2−2aex+a2x^2-2aex+a^2e^2+y^2=e^2x^2-2aex+a^2
    Expand: the −2aex-2aex terms match.
  3. x2(1−e2)+y2=a2(1−e2)x^2(1-e^2)+y^2=a^2(1-e^2)
    Collect terms.
  4. x2a2+y2a2(1−e2)=1\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1
    Divide by a2(1−e2)a^2(1-e^2).
So b2=a2(1−e2)b^2=a^2(1-e^2). The same algebra with e>1e>1 gives the hyperbola.
foci (±ae,0),directrices x=±ae\text{foci }(\pm ae,0),\quad\text{directrices }x=\pm\frac ae
where
ee
eccentricity.
aa
semi-major axis (ellipse) or half the vertex distance (hyperbola).
Worked exampleEccentricity and foci

For x225+y29=1\frac{x^2}{25}+\frac{y^2}9=1 find ee, the foci and the directrices.

Goal
Three facts.
Strategy
Use b2=a2(1−e2)b^2=a^2(1-e^2).
  1. a=5, b=3a=5,\ b=3
    Read off the denominators.
  2. 9=25(1−e2)⇒e2=16259=25(1-e^2)\Rightarrow e^2=\frac{16}{25}
    Substitute.
  3. e=45, ae=4, ae=254e=\frac45,\ ae=4,\ \frac ae=\frac{25}4
    Compute.
Answere=45e=\frac45; foci (±4,0)(\pm4,0); directrices x=±254x=\pm\frac{25}4.
✓ CheckAt the vertex (5,0)(5,0): PS=1PS=1, PM=254−5=54PM=\frac{25}4-5=\frac54, ratio 45\frac45 ✓.

2. Parameters, tangents and asymptotes

ellipse (acos⁡θ,bsin⁡θ);hyperbola (asec⁡θ,btan⁡θ)\text{ellipse }(a\cos\theta,b\sin\theta);\quad\text{hyperbola }(a\sec\theta,b\tan\theta)
where
θ\theta
parameter (not the polar angle of the point).
Tangent to the ellipse at (x1,y1)(x_1,y_1)
  1. 2xa2+2yb2dydx=0\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0
    Differentiate implicitly.
  2. dydx=−b2x1a2y1\frac{dy}{dx}=-\frac{b^2x_1}{a^2y_1}
    Rearrange at the point.
  3. xx1a2+yy1b2=x12a2+y12b2\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}
    Line through (x1,y1)(x_1,y_1), multiplied out.
  4. xx1a2+yy1b2=1\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1
    The point lies on the ellipse.
Worked exampleAn asymptote

Find the asymptotes of x216−y29=1\frac{x^2}{16}-\frac{y^2}9=1.

Goal
Two lines.
Strategy
For large xx the 11 is negligible.
  1. x216≈y29\frac{x^2}{16}\approx\frac{y^2}9
    Ignore the 11 far from the origin.
  2. y=±34xy=\pm\frac34x
    Square root.
Answery=±34xy=\pm\frac34x.
✓ Checkba=34\frac ba=\frac34 ✓.
Which conic?

The eccentricity alone decides the type of conic.

Common misconceptionVertices vs foci (±a,0)(\pm a,0) are vertices; the foci are (±ae,0)(\pm ae,0), inside the ellipse but outside the hyperbola's vertices.
Key takeawayFocal sum For an ellipse PS+PS′=2aPS+PS'=2a at every point: the string-and-pins property.
Pause and think

What is ee for x29−y216=1\frac{x^2}9-\frac{y^2}{16}=1?

Summary

  • Ellipse b2=a2(1−e2)b^2=a^2(1-e^2); hyperbola b2=a2(e2−1)b^2=a^2(e^2-1).
  • Foci (±ae,0)(\pm ae,0); directrices x=±aex=\pm\frac ae.
  • Tangent: replace x2→xx1x^2\to xx_1, y2→yy1y^2\to yy_1.
  • Asymptotes y=±baxy=\pm\frac bax.
Key words
Eccentricity
Ratio of distance to focus over distance to directrix.
Asymptote
Line a curve approaches but never reaches.
HistoryKepler. Johannes Kepler discovered in 1609 that planets move in ellipses with the Sun at one focus, overturning 2000 years of perfect circles.
Fun factKidney stones. Lithotripters put a shock-wave source at one focus of an elliptical reflector and the kidney stone at the other.
Section 3 of 14

Key formulae

Ellipsex2a2+y2b2=1,b2=a2(1−e2),(acos⁡θ,bsin⁡θ)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad b^2=a^2(1-e^2),\quad(a\cos\theta,b\sin\theta)
Hyperbolax2a2−y2b2=1,b2=a2(e2−1),y=±bax\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\quad b^2=a^2(e^2-1),\quad y=\pm\frac bax
Foci and directrices(±ae,0),x=±ae(\pm ae,0),\quad x=\pm\frac ae
TrickWhich conic? e<1e<1 ellipse, e=1e=1 parabola, e>1e>1 hyperbola. A plus sign between the squares means ellipse; minus means hyperbola.
Section 4 of 14

Worked examples

Example 1 · Ellipse0 / 3 steps
For x225+y29=1\frac{x^2}{25}+\frac{y^2}9=1 find ee, the foci and directrices.
    Example 2 · Tangent0 / 2 steps
    Find the tangent to x225+y29=1\frac{x^2}{25}+\frac{y^2}9=1 at (3,125)(3,\frac{12}5).
      Section 5 of 14

      Explore the graph

      Ellipse with its foci
      Loading the interactive graph…
      Try this
      1. Foci at (±ae,0)(\pm ae,0) with ae=a2−b2ae=\sqrt{a^2-b^2}.
      2. Make bb close to aa: the foci merge and the ellipse becomes a circle (e→0e\to0).

      Sum of focal distances =2a=2a everywhere.

      In the real worldOrbits. Halley's comet has e≈0.97e\approx0.97: a long thin ellipse. Objects on hyperbolic paths (e>1e>1), like the interstellar visitor ʻOumuamua, pass by once and never return.
      Section 6 of 14

      Common mistakes

      Incorrectb2=a2(1−e2)b^2=a^2(1-e^2) for a hyperbola
      Correcta2(e2−1)a^2(e^2-1)
      Hyperbolas have e>1e>1.
      IncorrectFoci at (±a,0)(\pm a,0)
      Correct(±ae,0)(\pm ae,0)
      (±a,0)(\pm a,0) are the vertices.
      IncorrectDirectrix x=aex=ae
      Correctx=aex=\frac ae
      Directrices lie outside the ellipse.

      Circle: e = 0

      Comet: e = 0.97, see you in 76 years.

      Section 7 of 14

      Exam tip

      Exam tipTangent shortcut. Replace x2x^2 by xx1xx_1 and y2y^2 by yy1yy_1 in the equation to get the tangent at (x1,y1)(x_1,y_1) instantly.
      Section 8 of 14

      Guided practice

      SA-FP-CE-G012 marksL2 StandardNon-calculatorEccentricity
      Find the eccentricity of x225+y29=1\frac{x^2}{25}+\frac{y^2}9=1.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-FP-CE-0011 markL1 FoundationNon-calculatorEccentricity
      For the same ellipse, find the positive xx-coordinate of a focus.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-CE-0021 markL2 StandardNon-calculatorEccentricity
      For the same ellipse, find the positive directrix x=kx=k.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-CE-0032 marksL2 StandardNon-calculatorParametric forms
      Find the positive gradient of the asymptotes of x216−y29=1\frac{x^2}{16}-\frac{y^2}9=1.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-CE-0043 marksL3 ExamNon-calculatorTangents
      Find the xx-intercept of the tangent to x225+y29=1\frac{x^2}{25}+\frac{y^2}9=1 at (3,125)\left(3,\frac{12}5\right).

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-FP-CE-0052 marksL3 ExamNon-calculatorEccentricity
      A point PP moves on x225+y29=1\frac{x^2}{25}+\frac{y^2}9=1. Find the sum of its distances to the two foci.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • Ellipse b2=a2(1−e2)b^2=a^2(1-e^2); hyperbola b2=a2(e2−1)b^2=a^2(e^2-1).
      • Foci (±ae,0)(\pm ae,0); directrices x=±aex=\pm\frac ae.
      • Hyperbola asymptotes y=±baxy=\pm\frac bax.
      Section 12 of 14

      Quick quiz

      Q1 · SA-FP-CE-Q011 markL1 FoundationNon-calculatorEccentricity
      A conic with eccentricity 1.51.5 is
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      IAL Further Mathematics (YFM01) and A level Further Mathematics (9FM0)
      Where
      IAL WFM03 Further coordinate systems; A level FP1 Conic sections
      Expected
      Ellipse and hyperbola in standard form; eccentricity, foci, directrices; parametric forms; tangents and normals; loci.
      Calculator
      Permitted; show enough working to earn the method marks.
      Typical style
      Edexcel often opens with a routine part and builds to a "show that" or "prove" part.
      Open in
      YFM01 · WFM039FM0 · Option
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

      Browse past papers

      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel 9FM0, First teaching 2017 (official specification).