AQA Further AS Paper 1 2020 June — Question 17 4 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2020
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeShow polar curve has Cartesian form
DifficultyStandard +0.8 This requires converting a polar equation to Cartesian form to identify the circle's radius, involving algebraic manipulation and completing the square. While the technique is standard for Further Maths students, the non-trivial polar form and need for full justification elevate it above routine exercises, placing it moderately above average difficulty.
Spec4.09a Polar coordinates: convert to/from cartesian

The polar equation of the circle \(C\) is $$r = a(\cos \theta + \sin \theta)$$ Find, in terms of \(a\), the radius of \(C\). Fully justify your answer. [4 marks]

Question 17:
AnswerMarks
17Selects a method to transform the given equation of C
into a standard polar form or Cartesian form, e.g.
uses and and ;
or 2 2 2
AnswerMarks Guidance
writes π‘Ÿπ‘Ÿ in= tπ‘₯π‘₯he +fo𝑦𝑦rm π‘₯π‘₯ = π‘Ÿπ‘Ÿcosπœƒπœƒ 𝑦𝑦 = π‘Ÿπ‘Ÿ. sinπœƒπœƒ3.1a M1
π‘Ÿπ‘Ÿ = π‘Žπ‘Ž(π‘Ÿπ‘Ÿcosπœƒπœƒ+π‘Ÿπ‘Ÿsin πœƒπœƒ)
2 2
π‘₯π‘₯ +𝑦𝑦 = π‘Žπ‘Ž(π‘₯π‘₯+𝑦𝑦)
2 2
π‘₯π‘₯ βˆ’π‘Žπ‘Žπ‘₯π‘₯+𝑦𝑦 βˆ’π‘Žπ‘Žπ‘¦π‘¦ = 0
2 2 2
2 π‘Žπ‘Ž 2 π‘Žπ‘Ž π‘Žπ‘Ž
π‘₯π‘₯ βˆ’π‘Žπ‘Žπ‘₯π‘₯+ +𝑦𝑦 βˆ’π‘Žπ‘Žπ‘¦π‘¦+ =
4 4 2
2 2 2
π‘Žπ‘Ž π‘Žπ‘Ž π‘Žπ‘Ž
οΏ½π‘₯π‘₯βˆ’2οΏ½ +οΏ½π‘¦π‘¦βˆ’2οΏ½ = �√2οΏ½
radius
π‘Žπ‘Ž
= √2
π‘Ÿπ‘Ÿ 𝑅𝑅(cos𝐴𝐴cos𝐡𝐡+sin𝐴𝐴sin𝐡𝐡)
Obtains a correct equation in terms of and only
or
obtains π‘₯π‘₯ 𝑦𝑦
.
πœ‹πœ‹
AnswerMarks Guidance
41.1b A1
π‘Ÿπ‘Ÿ = π‘Žπ‘Žβˆš2cosοΏ½πœƒπœƒβˆ’ οΏ½
Correctly completes the square of their quadratic
expression
or
states that the circle must pass through , and that the
maximum value of is 1.
πœ‹πœ‹ 𝑂𝑂
AnswerMarks Guidance
41.1a M1
cosοΏ½πœƒπœƒβˆ’ οΏ½
Obtains the correct radius = .
AnswerMarks Guidance
π‘Žπ‘Ž3.2a A1
√2
AnswerMarks Guidance
Total4
QMarking instructions AO
Question 17:
17 | Selects a method to transform the given equation of C
into a standard polar form or Cartesian form, e.g.
uses and and ;
or 2 2 2
writes π‘Ÿπ‘Ÿ in= tπ‘₯π‘₯he +fo𝑦𝑦rm π‘₯π‘₯ = π‘Ÿπ‘Ÿcosπœƒπœƒ 𝑦𝑦 = π‘Ÿπ‘Ÿ. sinπœƒπœƒ | 3.1a | M1 | 2
π‘Ÿπ‘Ÿ = π‘Žπ‘Ž(π‘Ÿπ‘Ÿcosπœƒπœƒ+π‘Ÿπ‘Ÿsin πœƒπœƒ)
2 2
π‘₯π‘₯ +𝑦𝑦 = π‘Žπ‘Ž(π‘₯π‘₯+𝑦𝑦)
2 2
π‘₯π‘₯ βˆ’π‘Žπ‘Žπ‘₯π‘₯+𝑦𝑦 βˆ’π‘Žπ‘Žπ‘¦π‘¦ = 0
2 2 2
2 π‘Žπ‘Ž 2 π‘Žπ‘Ž π‘Žπ‘Ž
π‘₯π‘₯ βˆ’π‘Žπ‘Žπ‘₯π‘₯+ +𝑦𝑦 βˆ’π‘Žπ‘Žπ‘¦π‘¦+ =
4 4 2
2 2 2
π‘Žπ‘Ž π‘Žπ‘Ž π‘Žπ‘Ž
οΏ½π‘₯π‘₯βˆ’2οΏ½ +οΏ½π‘¦π‘¦βˆ’2οΏ½ = �√2οΏ½
radius
π‘Žπ‘Ž
= √2
π‘Ÿπ‘Ÿ 𝑅𝑅(cos𝐴𝐴cos𝐡𝐡+sin𝐴𝐴sin𝐡𝐡)
Obtains a correct equation in terms of and only
or
obtains π‘₯π‘₯ 𝑦𝑦
.
πœ‹πœ‹
4 | 1.1b | A1
π‘Ÿπ‘Ÿ = π‘Žπ‘Žβˆš2cosοΏ½πœƒπœƒβˆ’ οΏ½
Correctly completes the square of their quadratic
expression
or
states that the circle must pass through , and that the
maximum value of is 1.
πœ‹πœ‹ 𝑂𝑂
4 | 1.1a | M1
cosοΏ½πœƒπœƒβˆ’ οΏ½
Obtains the correct radius = .
π‘Žπ‘Ž | 3.2a | A1
√2
Total | 4
Q | Marking instructions | AO | Marks | Typical solution
The polar equation of the circle $C$ is
$$r = a(\cos \theta + \sin \theta)$$

Find, in terms of $a$, the radius of $C$.

Fully justify your answer.
[4 marks]

\hfill \mbox{\textit{AQA Further AS Paper 1 2020 Q17 [4]}}