A LevelFurther MathsEdexcelFurther Pure 1Revision NotesCoordinate SystemsProperties of Ellipses, Parabolas & HyperbolasProperties of Ellipses

Properties of Ellipses (Edexcel A Level Further Maths): Revision Note

Exam code: 9FM0

Mark Curtis

Last updated

Properties of ellipses

What is an ellipse?

  • An ellipse is a stretched circle

    • a is its horizontal half length

    • b is its vertical half length

Three shapes: a horizontal ellipse with a greater width than height labelled 'a > b,' a circle labeled 'a = b,' and a vertical ellipse labelled 'a < b.'

  • If a greater than b

    • the major axis is negative a less or equal than x less or equal than a

      • of length 2 a

    • the minor axis is negative b less or equal than y less or equal than b

      • of length 2 b

    • 0 less or equal than x less or equal than a is called a semi-major axis

      • of length a

    • 0 less or equal than y less or equal than b is called a semi-minor axis

      • of length b

Diagram of an ellipse showing major axis as 2a, minor axis as 2b, semi-major axis as a, semi-minor axis as b, with a greater than b.

  • An ellipse is one of the conic curves

    • with eccentricity 0 less or equal than e less than 1

    • A circle is an ellipse with e equals 0

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

What is the equation of an ellipse?

  • The Cartesian equation of an ellipse with its centre at the origin

    • is x squared over a squared plus y squared over b squared equals 1

    • where

      • negative a less or equal than x less or equal than a

      • negative b less or equal than y less or equal than b

  • The parametric equations of an ellipse

    • are

      • x equals a cos theta

      • y equals b sin theta

    • where 0 less or equal than theta less than 2 pi

  • Eliminating the parameter, theta, gives the Cartesian equation x squared over a squared plus y squared over b squared equals 1

    • using that cos squared theta plus sin squared theta identical to 1

Examiner Tips and Tricks

You are given the Cartesian and parametric equations of an ellipse in the formulae booklet.

What are the coordinates of a general point on an ellipse?

  • A general point P on the ellipse x squared over a squared plus y squared over b squared equals 1 has coordinates given by its parametric equations, P open parentheses a cos theta comma space b sin theta close parentheses

Diagram of an ellipse with equation x^2/a^2 + y^2/b^2 = 1; axes labelled from -a to a horizontally and -b to b vertically with a general point P with coordinates (a cost, b sin t).

  • e.g. P open parentheses 3 cos theta comma space 2 sin theta close parentheses is a general point on the ellipse x squared over 9 plus y squared over 4 equals 1

    • It satisfies the equation of the curve

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

  • This is different to, say, open parentheses 3 comma space 0 close parentheses

    • which is a fixed point on the ellipse x squared over 9 plus y squared over 4 equals 1

What is the eccentricity, focus and directrix of an ellipse?

  • The eccentricity of an ellipse, e, where 0 less or equal than e less than 1, is a measure of how stretched the ellipse is

    • e equals 0 gives a perfect circle

    • e rightwards arrow 1 gets flatter and flatter

  • If a greater than b

    • the eccentricity is found be rearranging the following formula

      • b squared equals a squared open parentheses 1 minus e squared close parentheses

    • the foci, F and F apostrophe, are two symmetric points inside the ellipse on the major axis

      • with coordinates open parentheses plus-or-minus a e comma space 0 close parentheses

    • the directrices are the two vertical lines positioned symmetrically outside of the ellipse

      • with equations x equals plus-or-minus a over e

Diagram of an ellipse where a>b with foci F(ae, 0) and F'(-ae,0), centre O, semi-major axis a, semi-minor axis b, and eccentricity given by b^2=a^2(1-e^2). Two vertical lines (directrices) are at x=-a/e and x=a/e.

Examiner Tips and Tricks

You are given the eccentricity formula, foci and directrices of an ellipse in the formulae booklet.

  • If a less than b

    • the eccentricity is found be rearranging the following formula

      • a squared equals b squared open parentheses 1 minus e squared close parentheses

    • the foci, F and F apostrophe, are the symmetric points on the major axis

      • with coordinates open parentheses 0 comma space plus-or-minus b e close parentheses

    • the directrices are the two symmetric horizontal lines

      • with equations y equals plus-or-minus b over e

Diagram of an ellipse where a<b with foci F(0, be) and F'(0. -be), centre O, semi-major axis b, semi-minor axis a, and eccentricity given by a^2=b^2(1-e^2). Two horizontal lines (directrices) are at y=-b/e and y=b/e.

Examiner Tips and Tricks

You are not given any formulae for the a less than b case, but you can work them out by swapping 'horizontal to vertical' and ' a to b'.

Worked Example

An ellipse has the equation x squared over 25 plus y squared over 16 equals 1.

(a) Calculate the coordinates of the foci.

(b) Calculate the equations of the directrices.

Answer:

(a)

Find a and b by comparing to the general equation x squared over a squared plus y squared over b squared equals 1

a equals 5 b equals 4

Check that a greater than b

5 greater than 4

Rearrange the relationship b squared equals a squared open parentheses 1 minus e squared close parentheses to find e

  • and check that 0 less or equal than e less than 1

table row cell 4 squared end cell equals cell 5 squared open parentheses 1 minus e squared close parentheses end cell row cell 16 over 25 end cell equals cell 1 minus e squared end cell row cell e squared end cell equals cell 1 minus 16 over 25 end cell row cell e squared end cell equals cell 9 over 25 end cell row e equals cell 3 over 5 end cell end table

Calculate the foci using open parentheses plus-or-minus a e comma space 0 close parentheses

open parentheses plus-or-minus 5 cross times 3 over 5 comma space 0 close parentheses

The foci have coordinates open parentheses plus-or-minus 3 comma space 0 close parentheses

(b)

Calculate the equations of the directrices using x equals plus-or-minus a over e

x equals plus-or-minus fraction numerator 5 over denominator open parentheses 3 over 5 close parentheses end fraction

The directrices have equations x equals plus-or-minus 25 over 3

What is the focus-directrix property of an ellipse?

  • The focus-directrix property says that, if you take any point P on an ellipse, 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. fraction numerator P F over denominator P D end fraction equals e

      • sometimes rearranged to P F equals e P D

The focus-directrix property showing a general point P, the length PF from P to the focus at (ae, 0) and the length PD which is a horizontal distance from P to the directrix x=a/e. The formula is PF/PD = e.

  • The focus-directrix property works from P to the other focus, F apostrophe, and directrix, D apostrophe

    • fraction numerator P F apostrophe over denominator P D apostrophe end fraction equals 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

An ellipse with foci F and F apostrophe and directrices x equals plus-or-minus a over e is shown below.

The point P on the ellipse has coordinates open parentheses x comma space y close parentheses and the points D and D apostrophe are on the directrices, at the same height as P.

An ellipse between -a and a with the foci F and F' shown and vertical lines at x=-a/e and x=a/e. The point P lies on the ellipse in the first quadrant and has coordinates (x, y). The points D and D' are on the directrices at the same vertical height as P.

Using only the focus-directrix property,

(a) prove that P F apostrophe plus P F equals 2 a

(b) derive the Cartesian equation of an ellipse, x squared over a squared plus y squared over b squared equals 1, where b squared equals a squared open parentheses 1 minus e squared close parentheses

Answer:

(a)

Use the focus-directrix property on P, F apostrophe and D apostrophe and again on P, F and D

fraction numerator P F apostrophe over denominator P D apostrophe end fraction equals e space space and space space fraction numerator P F over denominator P D end fraction equals e

An ellipse with foci F, F' and directrices x=a/e and x=-a/e. The point P(x,y) is on the ellipse and the points D and D' are on the directrices at the same height. The lines PF, PF', PD and PD' are shown. Two formulas are shown: PF'/PD'=e and PF/PD=e. The total distance between the directrices is 2a/e.

Rearrange to make P F apostrophe and P F the subjects

P F apostrophe equals e P D apostrophe space space and space space P F equals e P D

Add together P F apostrophe and P F

P F apostrophe plus P F equals e P D apostrophe plus e P D

Factorise out e

P F apostrophe plus P F equals e open parentheses P D apostrophe plus P D close parentheses

Use that P D apostrophe plus P D is the total distance between the two directrices x equals plus-or-minus a over e

P D apostrophe plus P D equals a over e minus open parentheses negative a over e close parentheses equals fraction numerator 2 a over denominator e end fraction

Substitute this back into P F apostrophe plus P F

P F apostrophe plus P F equals e open parentheses fraction numerator 2 a over denominator e end fraction close parentheses

Simplify

P F apostrophe plus P F equals 2 a

(b)

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

fraction numerator P F over denominator P D end fraction equals e

It helps to draw lengths x and y from P open parentheses x comma space y close parentheses on the diagram and the foci open parentheses plus-or-minus a e comma space 0 close parentheses

Create a right-angled triangle whose hypotenuse is P F with base open parentheses a e minus x close parentheses and height y

An ellipse with foci F, F' and directrices x=a/e and x=-a/e. The point P(x,y) is on the ellipse and the points D and D' are on the directrices at the same height. The lines PF and PD are shown. The formula PF/PD=e is shown. A right-angled triangle has hypotenuse PF and base (ae-x) and height y. The distance from P to D is (a/e-x).

Use Pythagoras' theorem to find P F squared

P F squared equals open parentheses a e minus x close parentheses squared plus y squared

Find the length P D from x to the directrix

P D equals open parentheses a over e minus x close parentheses

Rearrange fraction numerator P F over denominator P D end fraction equals e to make P F squared the subject

P F squared equals e squared P D squared

Substitute in expressions for P F squared and P D squared from above

open parentheses a e minus x close parentheses squared plus y squared equals e squared open parentheses a over e minus x close parentheses squared

Expand, cancel and factorise

table row cell a squared e squared minus 2 a e x plus x squared plus y squared end cell equals cell e squared open parentheses a squared over e squared minus fraction numerator 2 a x over denominator e end fraction plus x squared close parentheses end cell row cell a squared e squared up diagonal strike negative 2 a e x end strike plus x squared plus y squared end cell equals cell a squared up diagonal strike negative 2 a e x end strike plus e squared x squared end cell row cell open parentheses 1 minus e squared close parentheses x squared plus y squared end cell equals cell a squared open parentheses 1 minus e squared close parentheses end cell end table

Divide both sides by a squared open parentheses 1 minus e squared close parentheses

x squared over a squared plus fraction numerator y squared over denominator a squared open parentheses 1 minus e squared close parentheses end fraction equals 1

This is now in the correct form of an ellipse

The Cartesian equation is x squared over a squared plus y squared over b squared equals 1, where b squared equals a squared open parentheses 1 minus e squared close parentheses

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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.