AQA FP3 2007 January — Question 2 6 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2007
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeConvert Cartesian to polar equation
DifficultyStandard +0.8 This requires systematic manipulation of polar-to-Cartesian conversions (r²=x²+y², y=r sin θ) and algebraic rearrangement to isolate y, including dealing with a square root. While the steps are methodical, the algebraic manipulation to reach explicit form y=f(x) requires careful handling and is more involved than routine conversion exercises.
Spec4.09a Polar coordinates: convert to/from cartesian
2 A curve has polar equation \(r ( 1 - \sin \theta ) = 4\). Find its cartesian equation in the form \(y = \mathrm { f } ( x )\).
Question 2:
AnswerMarks Guidance
WorkingMarks Guidance
\(r - r\sin\theta = 4\)M1
\(r - y = 4\)B1 \(r\sin\theta = y\) stated or used
\(r = y + 4\)A1
\(x^2 + y^2 = (y+4)^2\)M1 \(r^2 = x^2 + y^2\) used
\(x^2 + y^2 = y^2 + 8y + 16\)A1F ft one slip
\(y = \dfrac{x^2 - 16}{8}\)A1 Total: 6 
## Question 2:

| Working | Marks | Guidance |
|---------|-------|----------|
| $r - r\sin\theta = 4$ | M1 | |
| $r - y = 4$ | B1 | $r\sin\theta = y$ stated or used |
| $r = y + 4$ | A1 | |
| $x^2 + y^2 = (y+4)^2$ | M1 | $r^2 = x^2 + y^2$ used |
| $x^2 + y^2 = y^2 + 8y + 16$ | A1F | ft one slip |
| $y = \dfrac{x^2 - 16}{8}$ | A1 | Total: 6 | |

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2 A curve has polar equation $r ( 1 - \sin \theta ) = 4$. Find its cartesian equation in the form $y = \mathrm { f } ( x )$.

\hfill \mbox{\textit{AQA FP3 2007 Q2 [6]}}
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Q1 9 Q2 6 Q3 9 Q4 8 Q5 12 Q6 16 Q7 15