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Tangents & Normals (Edexcel IGCSE Further Pure Maths)

Revision Note

Paul

Written by: Paul

Reviewed by: Dan Finlay

Tangents & Normals

What is a tangent?

  • At any point on the graph of a (non-linear) function

    • the tangent is the straight line that touches the graph at the point without cutting through it

    • Its gradient is given by the derivative of the function

Tangent to a curve

How do I find the equation of a tangent?

  • You need a point and the gradient to find the equation of a straight line 

    • The gradient of the tangent to the function y equals straight f open parentheses x close parentheses at the point left parenthesis x subscript 1 comma blank y subscript 1 right parenthesis is straight f to the power of apostrophe stretchy left parenthesis x subscript 1 stretchy right parenthesis

  • Therefore to find the equation of the tangent to the function size 16px y size 16px equals size 16px f size 16px left parenthesis size 16px x size 16px right parenthesis at the point left parenthesis x subscript 1 comma blank y subscript 1 right parenthesis

    • Find  straight f to the power of apostrophe open parentheses x close parentheses

    • Substitute x subscript 1 into straight f apostrophe open parentheses x close parentheses to find the gradient straight f to the power of apostrophe open parentheses x subscript 1 close parentheses,

    • Use the  y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses form of the line equation

      •  y minus y subscript 1 equals straight f to the power of apostrophe stretchy left parenthesis x subscript 1 stretchy right parenthesis left parenthesis x minus x subscript 1 right parenthesis

    • Rearrange the equation into whatever form the question requires

What is a normal?

  • At any point on the graph of a (non-linear) function

    • the normal is the straight line that passes through that point

    • and is perpendicular to the tangent

Normal to a curve

How do I find the equation of a normal?

  • You need a point and the gradient to find the equation of a straight line

    • The tangent and the normal are perpendicular

    • Therefore the gradient of the normal to the function y equals straight f open parentheses x close parentheses at the point left parenthesis x subscript 1 comma blank y subscript 1 right parenthesis is negative fraction numerator 1 over denominator straight f to the power of apostrophe stretchy left parenthesis x subscript 1 stretchy right parenthesis end fraction

  • To find the equation of the normal to the function y equals straight f left parenthesis x right parenthesis at the point size 16px left parenthesis size 16px x subscript 1 size 16px comma blank size 16px y subscript 1 size 16px right parenthesis

    • Find  straight f to the power of apostrophe open parentheses x close parentheses

    • Substitute x subscript 1 into straight f apostrophe open parentheses x close parentheses to find the gradient of the tangent straight f to the power of apostrophe open parentheses x subscript 1 close parentheses

    • Use that to find the gradient of the normal negative fraction numerator 1 over denominator straight f to the power of apostrophe open parentheses x subscript 1 close parentheses end fraction

    • Use the  y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses form of the line equation

      •  y minus y subscript 1 equals negative fraction numerator 1 over denominator straight f to the power of apostrophe stretchy left parenthesis x subscript 1 stretchy right parenthesis end fraction left parenthesis x minus x subscript 1 right parenthesis

    • Rearrange the equation into whatever form the question requires

Examiner Tips and Tricks

  • Make sure you are confident with finding the equation of a straight line

    • In particular when you know the gradient and one point on the line

    • This is an essential skill for finding tangents and normals

Worked Example

The function straight f left parenthesis x right parenthesis is defined by

 straight f stretchy left parenthesis x stretchy right parenthesis equals 2 x to the power of 4 plus 3 over x squared blank x not equal to 0

a) Find an equation for the tangent to the curve y equals straight f left parenthesis x right parenthesis at the point where x equals 1, giving your answer in the form y equals m x plus c.

Substitute x equals 1 into straight f open parentheses x close parentheses to find the y-coordinate of the point

straight f open parentheses 1 close parentheses equals 2 open parentheses 1 close parentheses to the power of 4 plus 3 over open parentheses 1 close parentheses squared equals 2 plus 3 equals 5

So the point in question is open parentheses 1 comma space 5 close parentheses

Now differentiate to find straight f to the power of apostrophe open parentheses x close parentheses, first using laws of indices to rewrite 3 over x squared as 3 x to the power of negative 2 end exponent


straight f open parentheses x close parentheses equals 2 x to the power of 4 plus 3 x to the power of negative 2 end exponent

table row cell straight f to the power of apostrophe open parentheses x close parentheses end cell equals cell 2 open parentheses 4 x to the power of 4 minus 3 end exponent close parentheses plus 3 open parentheses negative 2 x to the power of negative 2 minus 1 end exponent close parentheses end cell row blank equals cell 8 x cubed minus 6 x to the power of negative 3 end exponent end cell row blank equals cell 8 x cubed minus 6 over x cubed end cell end table


Substitute x equals 1 into straight f to the power of apostrophe open parentheses x close parentheses to find the gradient of the tangent at the point

straight f to the power of apostrophe open parentheses 1 close parentheses equals 8 open parentheses 1 close parentheses cubed minus 6 over open parentheses 1 close parentheses cubed equals 8 minus 6 equals 2


Now use  y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses  to find the equation of the line

Here m equals 2 and open parentheses x subscript 1 comma space y subscript 1 close parentheses equals open parentheses 1 comma space 5 close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell y minus 5 end cell equals cell 2 open parentheses x minus 1 close parentheses end cell row cell y minus 5 end cell equals cell 2 x minus 2 end cell row y equals cell 2 x plus 3 end cell end table


bold italic y bold equals bold 2 bold italic x bold plus bold 3

b) Find an equation for the normal to the curve y equals straight f open parentheses x close parentheses at the point where x equals 1, giving your answer in the form a x plus b y equals c, where a, b and c are integers.

The normal and tangent are perpendicular

So the gradient of the normal will be  negative fraction numerator 1 over denominator straight f to the power of apostrophe open parentheses 1 close parentheses end fraction

negative fraction numerator 1 over denominator straight f to the power of apostrophe open parentheses 1 close parentheses end fraction equals negative 1 half


Now use  y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses  to find the equation of the line

Here m equals negative 1 half and open parentheses x subscript 1 comma space y subscript 1 close parentheses equals open parentheses 1 comma space 5 close parentheses

y minus 5 equals negative 1 half open parentheses x minus 1 close parentheses y minus 5 equals negative 1 half x plus 1 half


Multiply both sides by 2 to get rid of the fractions

Then rearrange into the required form

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 y minus 10 end cell equals cell negative x plus 1 end cell row cell x plus 2 y end cell equals 11 end table


bold italic x bold plus bold 2 bold italic y bold equals bold 11

 

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    Author
    Reviewer
    Paul

    Author: Paul

    Expertise: Maths Content Creator (Previous)

    Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.