| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Convert Cartesian to polar equation |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring conversion from polar to Cartesian form and recognition of a conic section transformation. The algebraic manipulation is straightforward (multiply through, substitute r² = x² + y², x = r cos θ), and identifying the ellipse parameters from standard form is routine for FP3 students. The transformation recognition in part (b) is immediate once part (a) is complete. More challenging than average A-level due to being Further Maths content, but a standard textbook exercise within that context. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = \frac{10}{3-2\cos\theta}; \quad 3r - 2r\cos\theta = 10\) | ||
| \(3r - 2x = 10\) | M1 | \(r\cos\theta = x\) used at any stage |
| \(3r = 2x + 10\) | A1 | \(3r = 2x + 10\) or more suitable correct form to allow elimination of \(r\); or a correct Cartesian equation in ACF |
| \(9r^2 = (2x+10)^2\) | ||
| \(9(x^2 + y^2) = (2x+10)^2\) | M1 | \(r^2 = x^2 + y^2\) used to form a Cartesian equation |
| \(5x^2 + 9y^2 - 40x = 100\) | ||
| \(5(x-4)^2 + 9y^2 = 180\) | m1 | Completing the square \(5x^2 + 9y^2 \pm 40x = 5[(x \pm 4)^2 - 4^2] + 9y^2\) OE; this m1 cannot be awarded retrospectively |
| \(\frac{(x-4)^2}{36} + \frac{y^2}{20} = 1\) | ||
| \(c = 36, \quad d = 20\) | A1 | CSO; \(c = 36\), \(d = 20\) stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Translation \(\begin{bmatrix} 4 \\ 0 \end{bmatrix}\) | B1F | ft is only applied if m1 scored in (a). Must be 'translat…' and vector form. B0F if more than one transformation |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \frac{10}{3-2\cos\theta}; \quad 3r - 2r\cos\theta = 10$ | | |
| $3r - 2x = 10$ | M1 | $r\cos\theta = x$ **used** at any stage |
| $3r = 2x + 10$ | A1 | $3r = 2x + 10$ or more suitable correct form to allow elimination of $r$; or a correct Cartesian equation in ACF |
| $9r^2 = (2x+10)^2$ | | |
| $9(x^2 + y^2) = (2x+10)^2$ | M1 | $r^2 = x^2 + y^2$ **used** to form a Cartesian equation |
| $5x^2 + 9y^2 - 40x = 100$ | | |
| $5(x-4)^2 + 9y^2 = 180$ | m1 | Completing the square $5x^2 + 9y^2 \pm 40x = 5[(x \pm 4)^2 - 4^2] + 9y^2$ OE; this m1 cannot be awarded retrospectively |
| $\frac{(x-4)^2}{36} + \frac{y^2}{20} = 1$ | | |
| $c = 36, \quad d = 20$ | A1 | CSO; $c = 36$, $d = 20$ stated |
**Total: 5 marks**
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Translation $\begin{bmatrix} 4 \\ 0 \end{bmatrix}$ | B1F | ft is only applied if m1 scored in **(a)**. Must be 'translat…' and vector form. **B0F** if more than one transformation |
**Total: 1 mark**4
\begin{enumerate}[label=(\alph*)]
\item The curve with Cartesian equation $\frac { x ^ { 2 } } { c } + \frac { y ^ { 2 } } { d } = 1$ is mapped onto the curve with polar equation $r = \frac { 10 } { 3 - 2 \cos \theta }$ by a single geometrical transformation.
By writing the polar equation as a Cartesian equation in a suitable form, find the values of the constants $c$ and $d$.
\item Hence describe the geometrical transformation referred to in part (a).\\[0pt]
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2016 Q4 [6]}}