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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn7367Paper 2 · Compulsory contentFurther algebra and functionsEllipse and hyperbola

Further algebra and functions · Subtopic 4 of 6

The Ellipse and Hyperbola

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

Edexcel YFM01 WFM03Edexcel 9FM0 OptionAQA 7367 Paper 1AQA 7367 Paper 2≈ 50 minRevision 5 · updated 2026-11-04

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Section 1 of 14

Learning objectives

Find eccentricity, foci and directrices.
Use parametric forms $(a\cos\theta,b\sin\theta)$ and $(a\sec\theta,b\tan\theta)$ .
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 $=e\times$ distance to directrix. For $e<1$ you get the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $b^2=a^2(1-e^2)$ ; for $e>1$ the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $b^2=a^2(e^2-1)$ .

1. Finding the foci

Ellipse from the focus–directrix rule

$(x-ae)^2+y^2=e^2\left(x-\tfrac ae\right)^2$
$PS^2=e^2PM^2$ with focus $(ae,0)$ and directrix $x=\frac ae$ .
$x^2-2aex+a^2e^2+y^2=e^2x^2-2aex+a^2$
Expand: the $-2aex$ terms match.
$x^2(1-e^2)+y^2=a^2(1-e^2)$
Collect terms.
$\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1$
Divide by $a^2(1-e^2)$ .

b^2=a^2(1-e^2)

. The same algebra with

e>1

gives the hyperbola.

\text{foci }(\pm ae,0),\quad\text{directrices }x=\pm\frac ae

$e$: eccentricity.
$a$: semi-major axis (ellipse) or half the vertex distance (hyperbola).

Worked exampleEccentricity and foci

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

Goal: Three facts.
Strategy: Use $b^2=a^2(1-e^2)$ .

$a=5,\ b=3$
Read off the denominators.
$9=25(1-e^2)\Rightarrow e^2=\frac{16}{25}$
Substitute.
$e=\frac45,\ ae=4,\ \frac ae=\frac{25}4$
Compute.

Answer

e=\frac45

; foci

(\pm4,0)

; directrices

x=\pm\frac{25}4

✓ CheckAt the vertex

(5,0)

PS=1

PM=\frac{25}4-5=\frac54

, ratio

\frac45

✓.

2. Parameters, tangents and asymptotes

\text{ellipse }(a\cos\theta,b\sin\theta);\quad\text{hyperbola }(a\sec\theta,b\tan\theta)

$\theta$: parameter (not the polar angle of the point).

Tangent to the ellipse at

(x_1,y_1)

$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$
Differentiate implicitly.
$\frac{dy}{dx}=-\frac{b^2x_1}{a^2y_1}$
Rearrange at the point.
$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}$
Line through $(x_1,y_1)$ , multiplied out.
$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$
The point lies on the ellipse.

Worked exampleAn asymptote

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

Goal: Two lines.
Strategy: For large $x$ the $1$ is negligible.

$\frac{x^2}{16}\approx\frac{y^2}9$
Ignore the $1$ far from the origin.
$y=\pm\frac34x$
Square root.

Answer

y=\pm\frac34x

✓ Check

\frac ba=\frac34

✓.

Which conic?

Drawing diagram…

The eccentricity alone decides the type of conic.

Common misconceptionVertices vs foci

(\pm a,0)

are vertices; the foci are

(\pm ae,0)

, inside the ellipse but outside the hyperbola's vertices.

Key takeawayFocal sum For an ellipse

PS+PS'=2a

at every point: the string-and-pins property.

Pause and think

What is $e$ for $\frac{x^2}9-\frac{y^2}{16}=1$ ?

Conics

Curve	Equation	Parametric form
Parabola	$y^2=4ax$	$(at^2,2at)$
Ellipse	$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$	$(a\cos\theta,b\sin\theta)$
Hyperbola	$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$	$(a\cosh t,b\sinh t)$

Summary

Ellipse $b^2=a^2(1-e^2)$ ; hyperbola $b^2=a^2(e^2-1)$ .
Foci $(\pm ae,0)$ ; directrices $x=\pm\frac ae$ .
Tangent: replace $x^2\to xx_1$ , $y^2\to yy_1$ .
Asymptotes $y=\pm\frac bax$ .

Eccentricity: Ratio of distance to focus over distance to directrix.
Asymptote: Line a curve approaches but never reaches.

Also on your syllabus

Key takeawaySimple loci problems To find a locus, write the coordinates of the moving point in terms of a parameter, call them

(X,Y)

, and eliminate the parameter. Example: the tangent to

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

P(a\cos\theta,b\sin\theta)

meets the axes at

A

and

B

; the midpoint of

AB

\left(\frac{a}{2\cos\theta},\frac{b}{2\sin\theta}\right)

, so

\cos\theta=\frac a{2X}

\sin\theta=\frac b{2Y}

and the locus is

\frac{a^2}{4X^2}+\frac{b^2}{4Y^2}=1

Key takeawayTransforming conics Translating by

\binom pq

replaces

x

with

x-p

and

y

with

y-q

\frac{(x-p)^2}{a^2}+\frac{(y-q)^2}{b^2}=1

is an ellipse centred at

(p,q)

. A stretch by factor

k

parallel to the

x

-axis replaces

x

with

\frac xk

(so a circle

x^2+y^2=1

becomes the ellipse

\frac{x^2}{k^2}+y^2=1

). Reflecting in the

x

-axis replaces

y

with

-y

; in the

y

-axis,

x

with

-x

; and in

y=x

swaps

x

and

y

(so

y^2=4ax

becomes

x^2=4ay

). Transform the asymptotes and foci in the same way.

Normals: the normal at a point is perpendicular to the tangent there, so its gradient is $-\frac1m$ where $m$ is the tangent gradient, found by implicit or parametric differentiation.

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

Ellipse

\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)

Hyperbola

\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

(\pm ae,0),\quad x=\pm\frac ae

TrickWhich conic?

e<1

ellipse,

e=1

parabola,

e>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

\frac{x^2}{25}+\frac{y^2}9=1

find

e

, the foci and directrices.

Example 2 · Tangent0 / 2 steps

Find the tangent to

\frac{x^2}{25}+\frac{y^2}9=1

(3,\frac{12}5)

Section 5 of 14

Explore the graph

Ellipse with its foci

Loading the interactive graph…

Try this

Foci at $(\pm ae,0)$ with $ae=\sqrt{a^2-b^2}$ .
Make $b$ close to $a$ : the foci merge and the ellipse becomes a circle ( $e\to0$ ).

Sum of focal distances $=2a$ everywhere.

In the real worldOrbits. Halley's comet has

e\approx0.97

: a long thin ellipse. Objects on hyperbolic paths (

e>1

), like the interstellar visitor ʻOumuamua, pass by once and never return.

Section 6 of 14

Common mistakes

Incorrect

b^2=a^2(1-e^2)

for a hyperbola

Correct

a^2(e^2-1)

Hyperbolas have

e>1

IncorrectFoci at

(\pm a,0)

Correct

(\pm ae,0)

(\pm a,0)

are the vertices.

IncorrectDirectrix

x=ae

Correct

x=\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

x^2

xx_1

and

y^2

yy_1

in the equation to get the tangent at

(x_1,y_1)

instantly.

Section 8 of 14

Guided practice

SA-FP-CE-G012 marksL2 StandardNon-calculatorEccentricity

Find the eccentricity of

\frac{x^2}{25}+\frac{y^2}9=1

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

x

-coordinate of a focus.

Salam Academy Original Practice · Proposed mark scheme

SA-FP-CE-0021 markL2 StandardNon-calculatorEccentricity

For the same ellipse, find the positive directrix

x=k

Salam Academy Original Practice · Proposed mark scheme

SA-FP-CE-0032 marksL2 StandardNon-calculatorParametric forms

Find the positive gradient of the asymptotes of

\frac{x^2}{16}-\frac{y^2}9=1

Salam Academy Original Practice · Proposed mark scheme

SA-FP-CE-0043 marksL3 ExamNon-calculatorTangents

Find the

x

-intercept of the tangent to

\frac{x^2}{25}+\frac{y^2}9=1

\left(3,\frac{12}5\right)

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

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Section 11 of 14

Summary

Ellipse $b^2=a^2(1-e^2)$ ; hyperbola $b^2=a^2(e^2-1)$ .
Foci $(\pm ae,0)$ ; directrices $x=\pm\frac ae$ .
Hyperbola asymptotes $y=\pm\frac bax$ .

Section 12 of 14

Quick quiz

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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 · WFM03 9FM0 · Option

Section 14 of 14

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7367, First teaching 2017 (official specification).