All subjects · Further Mathematics · WFM03 Further Pure Mathematics 3

Foci, directrices and eccentricity

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

Every ellipse, parabola and hyperbola can be defined by one rule: the distance to a fixed point (the focus) is a fixed multiple e of the distance to a fixed line (the directrix). The number e is the eccentricity and decides which curve you get.

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The focus-directrix property

A point P moves so that PS = e × PM, where S is the focus and PM is the perpendicular distance to the directrix.

The closer e is to 0, the rounder the ellipse.

Ellipse x2/a2 + y2/b2 = 1 (a > b)

b2 = a2(1 - e2). Foci (±ae, 0). Directrices x = ±a/e.

For any point P on the ellipse, PS + PS' = 2a, where S and S' are the two foci. This is the 'pins and string' way to draw an ellipse.

Hyperbola x2/a2 - y2/b2 = 1

b2 = a2(e2 - 1). Foci (±ae, 0). Directrices x = ±a/e.

For any point P on the hyperbola, |PS - PS'| = 2a.

EllipseHyperbola
Eccentricity0 < e < 1e > 1
Link between a, b, eb2 = a2(1 - e2)b2 = a2(e2 - 1)
Foci(±ae, 0)(±ae, 0)
Directricesx = ±a/ex = ±a/e
Focal distancesPS + PS' = 2a|PS - PS'| = 2a

Worked example

Find the eccentricity, foci and directrices of x2/25 + y2/16 = 1.

  1. a = 5, b = 4, and 16 = 25(1 - e2), so e2 = 9/25 and e = 3/5
  2. ae = 3, so the foci are (±3, 0)
  3. a/e = 5 ÷ 3/5 = 25/3

Answer: e = 3/5, foci (±3, 0), directrices x = ±25/3

Key idea: PS = e PM. Ellipse: b2 = a2(1 - e2), e < 1. Hyperbola: b2 = a2(e2 - 1), e > 1. Foci (±ae, 0), directrices x = ±a/e. Ellipse PS + PS' = 2a.

Check yourself

  1. Find the eccentricity of the ellipse x2/25 + y2/16 = 1. Give your answer as a fraction in its simplest form.
    Show answer

    3/5. 16 = 25(1 - e2), so e2 = 9/25 and e = 3/5.

  2. For the ellipse x2/25 + y2/16 = 1, find the positive value of x for which x = k is a directrix. Give your answer as a fraction in its simplest form.
    Show answer

    25/3. x = a/e = 5 ÷ (3/5) = 25/3.

  3. Find the eccentricity of the hyperbola x2/9 - y2/16 = 1. Give your answer as a fraction in its simplest form.
    Show answer

    5/3. 16 = 9(e2 - 1), so e2 = 25/9 and e = 5/3.

  4. Find the positive x-coordinate of a focus of x2/9 - y2/16 = 1.
    Show answer

    5. ae = 3 × 5/3 = 5, so the foci are (±5, 0).

  5. An ellipse x2/a2 + y2/b2 = 1 has a = 4 and eccentricity 1/2. Find b2.
    Show answer

    12. b2 = a2(1 - e2) = 16(1 - 1/4) = 12.

  6. P lies on the ellipse x2/25 + y2/9 = 1 and is 3 units from one focus. How far is P from the other focus?
    Show answer

    7. PS + PS' = 2a = 10, so PS' = 10 - 3 = 7.

  7. A curve has focus-directrix property PS = 2PM. What is it?
    • An ellipse
    • A parabola
    • A hyperbola
    • A circle
    Show answer

    A hyperbola. e = 2 > 1 gives a hyperbola.

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Last updated 8 November 2026. Spotted a mistake? Tell us.