AQA FP3 2014 June — Question 3 4 marks

Exam BoardAQA
ModuleFP3 (Further Pure Mathematics 3)
Year2014
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeConvert Cartesian to polar equation
DifficultyStandard +0.8 This is a Further Maths polar coordinates question requiring manipulation of the conversion formulas (r² = x² + y², x = r cos θ) and algebraic rearrangement to eliminate r and θ. While the steps are systematic, it requires careful handling of substitutions and rearranging to the specific form requested, making it moderately challenging but still a standard FP3 exercise.
Spec4.09a Polar coordinates: convert to/from cartesian

3 A curve has polar equation \(r ( 4 - 3 \cos \theta ) = 4\). Find its Cartesian equation in the form \(y ^ { 2 } = \mathrm { f } ( x )\).
[0pt] [4 marks]

Question 3
[4 marks]
M1: Use \(r(4 - 3\cos\theta) = 4\) and \(x = r\cos\theta\)
M1: Rearrange to \(r = \frac{4}{4 - 3\cos\theta}\)
M1: Eliminate parameter using \(\cos\theta = \frac{x}{r}\) and \(r^2 = x^2 + y^2\)
A1: Correct Cartesian equation \(y^2 = f(x)\) in required form
## Question 3
[4 marks]

M1: Use $r(4 - 3\cos\theta) = 4$ and $x = r\cos\theta$

M1: Rearrange to $r = \frac{4}{4 - 3\cos\theta}$

M1: Eliminate parameter using $\cos\theta = \frac{x}{r}$ and $r^2 = x^2 + y^2$

A1: Correct Cartesian equation $y^2 = f(x)$ in required form
3 A curve has polar equation $r ( 4 - 3 \cos \theta ) = 4$. Find its Cartesian equation in the form $y ^ { 2 } = \mathrm { f } ( x )$.\\[0pt]
[4 marks]

\hfill \mbox{\textit{AQA FP3 2014 Q3 [4]}}