Properties of Hyperbolas (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Mark Curtis

Written by: Mark Curtis

Updated on

Properties of hyperbolas

What is a hyperbola?

  • A hyperbola is a curve with the Cartesian equation

    • x2a2y2b2=1

    • which consists of two separate branches

      • the positive-x branch (xa)

      • the negative-x branch (xa)

    • that are bounded by the two asymptotes

      • y=±bax

    • which pass through the origin

      • and are not necessarily perpendicular

Graph of a hyperbola with asymptotes y = (±b/a)x. Equation x²/a² - y²/b² = 1. Axes labelled with −a and a intercepts. The hyperbola is the shape of two branches (a C shape on the right and a backwards C shape on the left).
  • The asymptote equations can be derived by rearranging the curve

    • y2=b2x2a2(1a2x2) so y=±b2x2a21a2x2

      • as x, a2x20 so second square root tends to 1 and y±bax

  • A hyperbola is one of the conic curves

    • with eccentricity e>1

Diagram of conic sections showing circles, ellipses, parabolas, and hyperbolas formed by intersecting a plane with cones.

Examiner Tips and Tricks

You are given the Cartesian equation of a hyperbola and the equations of the asymptotes in the formulae booklet.

What are the parametric equations of a hyperbola?

  • There are two different sets of parametric equations for a hyperbola that are both equally valid

  • The first set of parametric equations is

    • x=±acoshθ

    • y=bsinhθ

    • where θ

      • and ± depends on positive or negative-x branch

  • The second set of parametric equations is

    • x=asecθ

    • y=btanθ

    • where π2<θ<π2 defines the positive-x branch

      • as secθ is positive

    • and both π<θ<π2 and π2<θπ define the negative-x branch

      • as secθ is negative

    • and θ±π2

      • as secθ and tanθ are undefined

  • Eliminating the parameter, θ, gives the Cartesian equation x2a2y2b2=1

    • using either cosh2θsinh2θ1

    • or 1+tan2θsec2θ and rearranging

Examiner Tips and Tricks

You are given both sets of parametric equations of a hyperbola in the formulae booklet (but not the ranges of θ).

Examiner Tips and Tricks

In the exam, unless given (or seen in subsequent results), you can use either set of parametric equations for a hyperbola.

What are the coordinates of a general point on a hyperbola?

  • A general point P on the hyperbola x2a2y2b2=1 has coordinates given by its parametric equations, P(acoshθ, bsinhθ) or P(asecθ, btanθ)

Graph of a hyperbola with asymptotes, labelled with equations. Point P has either coordinates (a cosh theta, b sinh theta) or (a sec theta, b tan theta)
  • e.g. P(3secθ, 2tanθ) is a general point on the hyperbola x29y24=1

    • It satisfies the equation of the curve

    • It moves around the curve depending on the value of θ

  • This is different to, say, (3, 0)

    • which is a fixed point on the hyperbola x29y24=1

What is the eccentricity, focus and directrix of a hyperbola?

  • The eccentricity of a hyperbola, e, takes the range e>1

  • If a>b

    • the eccentricity is found be rearranging the following formula

      • b2=a2(e21)

    • the foci, F and F', are two symmetric points on the x-axis enclosed by either branch

      • with coordinates (±ae, 0)

    • the directrices are the two vertical lines positioned symmetrically either side of the origin

      • in the gap between the branches

      • with equations x=±ae

Diagram showing a hyperbola x^2/a^2-y^2/b^2=1 with asymptotes y=(b/a)x and y=-(b/a)x, labelled axes, and focal points F and F' at ae and -ae on the x-axis. Two vertical lines x=-a/e and x=a/e are shown (directrices).

Examiner Tips and Tricks

You are given the eccentricity formula, foci and directrices of a hyperbola in the formulae booklet.

Worked Example

A hyperbola has the equation x216y29=1.

Calculate

(a) the coordinates of the foci,

(b) the equations of the directrices,

(c) the equations of any asymptotes.

Answer:

(a)

Find a and b by comparing to the general equation x2a2y2b2=1

a=4b=3

Check that a>b

4>3

Rearrange the relationship b2=a2(e21) to find e

  • and check that e>1

32=42(e21)916=e21e2=916+1e2=2516e=54

Calculate the foci using (±ae, 0)

(±4×54, 0)

The foci have coordinates (±5, 0)

(b)

Calculate the equations of the directrices using x=±ae

x=±4(54)

The directrices have equations x=±165

(c)

Substitute a=4 and b=3 into y=±bax

The asymptotes have equations y=±34x

What is the focus-directrix property of a hyperbola?

  • The focus-directrix property says that, if you take any point P on a hyperbola, then

    • the distance from P to the focus, F

    • divided by the shortest distance from P to the directrix (at point D)

    • is always equal to e, the eccentricity

    • i.e. PFPD=e

      • sometimes rearranged to PF=ePD

Diagram of a hyperbola with foci at F and F' at ae and -ae on the x-axis. A point P on the curve in the first quadrant is shown. The point D on the directrix x=a/e is shown at the same vertical height as P. Lines PD and PF are drawn. Formula for PF/PD = e.
  • The focus-directrix property works from P to the other focus, F', and directrix, D'

    • PF'PD'=e

      • where e is the same eccentricity

Examiner Tips and Tricks

You are not given the focus-directrix property in the exam (you must learn it).

Worked Example

One branch of a hyperbola with focus F at (ae, 0)and directrix x=ae is shown below.

The point P on the hyperbola has coordinates (x, y) and the point D is on the directrix, at the same height as P.

Diagram showing the right-hand branch of a hyperbola with the focus F at (ae, 0) and the directrix at x=a/e. A point  P(x,y) is labelled on the curve.

Using only the focus-directrix property, derive the Cartesian equation of a hyperbola, x2a2y2b2=1, where b2=a2(e21)

Answer:

Use the focus-directrix property on P, F and D (draw the lines PF and PD)

PFPD=e

It helps to draw the lengths x and y from P(x, y) on the diagram

Create a right-angled triangle whose hypotenuse is PF with base (xae) and height y

Diagram showing the right-hand branch of a hyperbola with the focus F at (ae, 0) and the directrix at x=a/e. A point  P(x,y) is labelled on the curve. PF is a right-angled triangle with base (x-ae) and height y. The distance between x and the directrix is (x-a/e). The formula PF/PD=e is shown.

Use Pythagoras' theorem to find PF2

PF2=(xae)2+y2

Find the length PD from x to the directrix

PD=(xae)

Rearrange PFPD=e to make PF2 the subject

PF2=e2PD2

Substitute in expressions for PF2 and PD2 from above

(xae)2+y2=e2(xae)2

Expand, cancel and factorise

x22aex+a2e2+y2=e2(x22axe+a2e2)x22aex+a2e2+y2=e2x22aex+a2a2(e21)=(e21)x2y2

Divide both sides by a2(e21)

1=x2a2y2a2(e21)

This now has the correct form of a hyperbola

The Cartesian equation is x2a2y2b2=1, where b2=a2(e21)

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Mark Curtis

Author: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.