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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

Learn9FM0Option · Further Pure Mathematics 1Conic sectionsParabola and rectangular hyperbola

Conic sections · Subtopic 1 of 2

The Parabola and Rectangular Hyperbola

The parabola $y^2=4ax$ is every point equidistant from the focus $(a,0)$ and the directrix $x=-a$ ; parametrically $(at^2,2at)$ . The rectangular hyperbola $xy=c^2$ has parametric form $\left(ct,\frac ct\right)$ . Parameters make tangents and normals tidy.

Edexcel YFM01 WFM01Edexcel 9FM0 Option≈ 45 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Use the focus–directrix property of $y^2=4ax$ .
Use the parametric forms $(at^2,2at)$ and $\left(ct,\frac ct\right)$ .
Find equations of tangents and normals.

Pause and think

Why are car headlight reflectors and satellite dishes parabolic?

Section 2 of 14

Core idea

Start hereWhy a satellite dish works

A satellite dish catches weak signals arriving in parallel and sends every one of them to a single receiver. Only one shape can do that: the parabola. Its secret is a simple distance rule, and that rule gives us its equation.

The big idea

A parabola is every point that is the same distance from a fixed point (the focus) and a fixed line (the directrix). With focus $(a,0)$ and directrix $x=-a$ this gives $y^2=4ax$ , which we can trace with the parameter $(at^2,2at)$ .

1. From the distance rule to the equation

Deriving

y^2=4ax

$PS=\sqrt{(x-a)^2+y^2}$
Distance from $P(x,y)$ to the focus $S(a,0)$ .
$PM=x+a$
Horizontal distance from $P$ to the line $x=-a$ .
$(x-a)^2+y^2=(x+a)^2$
Set $PS=PM$ and square both sides.
$x^2-2ax+a^2+y^2=x^2+2ax+a^2$
Expand both brackets.
$y^2=4ax$
Cancel $x^2$ and $a^2$ ; collect the $x$ terms.

Focus

(a,0)

, directrix

x=-a

y^2=4ax,\quad x=at^2,\ y=2at

$a$: distance from the vertex to the focus.
$t$: the parameter; each value gives one point.

2. Gradients without square roots

Tangent at

(at^2,2at)

$\frac{dx}{dt}=2at,\ \frac{dy}{dt}=2a$
Differentiate each coordinate.
$\frac{dy}{dx}=\frac{2a}{2at}=\frac1t$
Chain rule: divide the two rates.
$y-2at=\frac1t(x-at^2)$
Point–gradient form.
$ty=x+at^2$
Multiply by $t$ and tidy.

Normal gradient

-t

y+tx=2at+at^3

xy=c^2:\quad x=ct,\ y=\frac ct,\quad \frac{dy}{dx}=-\frac1{t^2}

$c$: scale; the curve passes through $(c,c)$ .

Worked exampleTangent to a parabola

Find the tangent to $y^2=12x$ at the point with $t=2$ .

Goal: An equation of a line.
Strategy: Identify $a$ , find the point, use gradient $\frac1t$ .

$4a=12\Rightarrow a=3$
Compare with $y^2=4ax$ .
$(at^2,2at)=(12,12)$
Substitute $t=2$ .
$m=\frac1t=\frac12$
Tangent gradient.
$y-12=\frac12(x-12)\Rightarrow2y=x+12$
Point–gradient form.

Answer

2y=x+12

✓ CheckThe formula

ty=x+at^2

gives

2y=x+12

✓.

Worked exampleTangent to a rectangular hyperbola

Find the tangent to $xy=9$ at $(3,3)$ .

Goal: A line.
Strategy: Use $c=3$ , $t=1$ .

$c=3,\ (ct,\tfrac ct)=(3,3)\Rightarrow t=1$
Match the point.
$m=-\frac1{t^2}=-1$
Gradient from the parameters.
$y-3=-(x-3)$
Point–gradient form.

Answer

x+y=6

✓ CheckIt touches

xy=9

only at

(3,3)

x(6-x)=9\Rightarrow(x-3)^2=0

✓.

Reflective property

Drawing diagram…

A ray parallel to the axis always reflects through the focus.

Common misconceptionReading off a For

y^2=20x

a=5

(not

20

). Always compare with

4a

Edge caseVertex

t=0

gives the vertex

(0,0)

, where the tangent is vertical (

x=0

) and

\frac1t

is undefined.

Pause and think

Where is the focus of $y^2=x$ ?

Summary

$y^2=4ax$ : focus $(a,0)$ , directrix $x=-a$ .
Parameter $(at^2,2at)$ , gradient $\frac1t$ .
$xy=c^2$ : $(ct,\frac ct)$ , gradient $-\frac1{t^2}$ .

Focus: Fixed point used to define a conic.
Directrix: Fixed line used to define a conic.
Rectangular hyperbola: Hyperbola with perpendicular asymptotes, e.g. $xy=c^2$ .

HistoryApollonius. Apollonius of Perga named the parabola, ellipse and hyperbola around 200 BC by slicing cones with planes at different angles.

Fun factWhispering dishes. Two parabolic dishes facing each other across a museum hall let a whisper at one focus be heard clearly at the other.

Section 3 of 14

Key formulae

Parabola

y^2=4ax:\ \text{focus }(a,0),\ \text{directrix }x=-a,\ (at^2,2at)

Rectangular hyperbola

xy=c^2:\ \left(ct,\tfrac ct\right),\ \text{asymptotes the axes}

Tangents

y^2=4ax:\ ty=x+at^2;\qquad xy=c^2:\ x+t^2y=2ct

TrickGradient from parameters.

\frac{dy}{dx}=\frac{dy/dt}{dx/dt}

: for

(at^2,2at)

that's

\frac{2a}{2at}=\frac1t

, so tangents and normals come out without square roots.

Section 4 of 14

Worked examples

Example 1 · Tangent to a parabola0 / 3 steps

Find the tangent to

y^2=12x

t=2

Example 2 · Rectangular hyperbola0 / 2 steps

Find the tangent to

xy=9

(3,3)

Section 5 of 14

Explore the graph

Focus and directrix

Loading the interactive graph…

Try this

The point $(at^2,2at)$ is $a(t^2+1)$ from the focus and from the directrix.
Move $t$ and check the two distances stay equal.

Distance to focus = distance to directrix.

In the real worldSolar furnaces. The Odeillo solar furnace in France uses a huge parabolic mirror to focus sunlight to over

3000^\circ

C at its focus.

Section 6 of 14

Common mistakes

IncorrectFocus of

y^2=12x

(12,0)

Correct

(3,0)

Compare with

4a=12

IncorrectNormal gradient

t

Correct

-t

Negative reciprocal of

\frac1t

Incorrect

xy=c^2

parameter

(t,\frac1t)

Correct

(ct,\frac ct)

Include

c

Every light ray parallel to the axis

Parabola: straight to the focus, all of you.

Section 7 of 14

Exam tip

Exam tipWork in t. Keep everything in terms of the parameter until the very end; it avoids messy square roots.

Section 8 of 14

Guided practice

SA-FP-CP-G011 markL1 FoundationNon-calculatorFocus–directrix

Find the

x

-coordinate of the focus of

y^2=12x

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FP-CP-0011 markL1 FoundationNon-calculatorFocus–directrix

The directrix of

y^2=8x

x=k

. Find

k

Salam Academy Original Practice · Proposed mark scheme

SA-FP-CP-0021 markL1 FoundationNon-calculatorParametric forms

For

y^2=8x

, find the

y

-coordinate of the point with parameter

t=3

Salam Academy Original Practice · Proposed mark scheme

SA-FP-CP-0032 marksL2 StandardNon-calculatorTangents

Find the gradient of the tangent to

y^2=4ax

at the point with parameter

t=2

Salam Academy Original Practice · Proposed mark scheme

SA-FP-CP-0041 markL1 FoundationNon-calculatorParametric forms

For

xy=16

with parametric form

(4t,\frac4t)

, find

y

when

t=2

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-CP-0053 marksL3 ExamNon-calculatorTangents

Find the

x

-intercept of the tangent to

xy=9

(3,3)

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

$y^2=4ax$ : focus $(a,0)$ , directrix $x=-a$ .
Parameters $(at^2,2at)$ , $(ct,\frac ct)$ .
Gradient $\frac1t$ on the parabola.

Section 12 of 14

Quick quiz

Q1 · SA-FP-CP-Q011 markL1 FoundationNon-calculatorFocus–directrix

A parabola is the set of points

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 WFM01 Coordinate systems; A level FP1 Conic sections
Expected: Parabola y² = 4ax with focus (a, 0) and directrix x = −a; rectangular hyperbola xy = c²; parametric forms (at², 2at) and (ct, c/t); tangents and normals.
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 · WFM01 9FM0 · Option

Section 14 of 14

Past paper questions

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