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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
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Learn7367Paper 2 · Compulsory contentFurther algebra and functionsParabola and rectangular hyperbola

Further algebra and functions · Subtopic 2 of 6

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 OptionAQA 7367 Paper 1AQA 7367 Paper 2≈ 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

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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

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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 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

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