| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (sin/cos identities) |
| Difficulty | Standard +0.8 This requires expanding a compound angle formula for cos(t + π/6), then manipulating two trigonometric parametric equations to eliminate the parameter t using the identity sin²t + cos²t = 1. The algebraic manipulation to reach the specific form (x+y)² + ay² = b is non-trivial and requires careful expansion and collection of terms. This goes beyond routine conversion and demands systematic algebraic skill, placing it moderately above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05l Double angle formulae: and compound angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x + y = 4\left(\cos t\cos\left(\frac{\pi}{6}\right) - \sin t\sin\left(\frac{\pi}{6}\right)\right) + 2\sin t\) | M1 | 3.1a — Looks ahead to the final result and uses the compound angle formula in a full attempt to write down an expression for \(x + y\) which is in terms of \(t\) only |
| Applies compound angle formula on their term in \(x\) | M1 | 1.1b — \(\cos\left(t + \frac{\pi}{6}\right) \rightarrow \cos t\cos\left(\frac{\pi}{6}\right) \pm \sin t\sin\left(\frac{\pi}{6}\right)\) |
| \(x + y = 2\sqrt{3}\cos t\) | A1 | 1.1b — Uses correct algebra to find \(x + y = 2\sqrt{3}\cos t\) |
| \(\left(\frac{x+y}{2\sqrt{3}}\right)^2 + \left(\frac{y}{2}\right)^2 = 1\) | M1 | 3.1a — Complete strategy of applying \(\cos^2 t + \sin^2 t = 1\) on rearranged \(x + y = "2\sqrt{3}\cos t"\), \(y = 2\sin t\) to achieve an equation in \(x\) and \(y\) only |
| \((x+y)^2 + 3y^2 = 12\) | A1 | 2.1 — Correctly proves \((x+y)^2 + ay^2 = b\) with both \(a=3\), \(b=12\), and no errors seen |
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x+y)^2 = \left(4\cos\left(t+\frac{\pi}{6}\right) + 2\sin t\right)^2\) | ||
| \(= \left(4\left(\cos t\cos\left(\frac{\pi}{6}\right) - \sin t\sin\left(\frac{\pi}{6}\right)\right) + 2\sin t\right)^2\) | M1 | 3.1a — Apply in the same way as main scheme |
| Applies compound angle formula | M1 | 1.1b — Apply in the same way as main scheme |
| \(= \left(2\sqrt{3}\cos t\right)^2\) or \(12\cos^2 t\) | A1 | 1.1b — Uses correct algebra to find \((x+y)^2 = \left(2\sqrt{3}\cos t\right)^2\) or \((x+y)^2 = 12\cos^2 t\) |
| \((x+y)^2 = 12(1-\sin^2 t) = 12 - 12\sin^2 t = 12 - 12\left(\frac{y}{2}\right)^2\) | M1 | 3.1a — Complete strategy of applying \(\cos^2 t + \sin^2 t = 1\) on \((x+y)^2 = \left("2\sqrt{3}\cos t"\right)^2\) to achieve an equation in \(x\) and \(y\) only |
| \((x+y)^2 + 3y^2 = 12\) | A1 | 2.1 — Correctly proves \((x+y)^2 + ay^2 = b\) with both \(a=3\), \(b=12\), and no errors seen |
| (5 marks) |
## Question 14:
**Main Scheme:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x + y = 4\left(\cos t\cos\left(\frac{\pi}{6}\right) - \sin t\sin\left(\frac{\pi}{6}\right)\right) + 2\sin t$ | M1 | 3.1a — Looks ahead to the final result and uses the compound angle formula in a full attempt to write down an expression for $x + y$ which is in terms of $t$ only |
| Applies compound angle formula on their term in $x$ | M1 | 1.1b — $\cos\left(t + \frac{\pi}{6}\right) \rightarrow \cos t\cos\left(\frac{\pi}{6}\right) \pm \sin t\sin\left(\frac{\pi}{6}\right)$ |
| $x + y = 2\sqrt{3}\cos t$ | A1 | 1.1b — Uses correct algebra to find $x + y = 2\sqrt{3}\cos t$ |
| $\left(\frac{x+y}{2\sqrt{3}}\right)^2 + \left(\frac{y}{2}\right)^2 = 1$ | M1 | 3.1a — Complete strategy of applying $\cos^2 t + \sin^2 t = 1$ on rearranged $x + y = "2\sqrt{3}\cos t"$, $y = 2\sin t$ to achieve an equation in $x$ and $y$ only |
| $(x+y)^2 + 3y^2 = 12$ | A1 | 2.1 — Correctly proves $(x+y)^2 + ay^2 = b$ with both $a=3$, $b=12$, and no errors seen |
| **(5 marks)** | | |
---
**Alternative Method (Alt 1):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x+y)^2 = \left(4\cos\left(t+\frac{\pi}{6}\right) + 2\sin t\right)^2$ | | |
| $= \left(4\left(\cos t\cos\left(\frac{\pi}{6}\right) - \sin t\sin\left(\frac{\pi}{6}\right)\right) + 2\sin t\right)^2$ | M1 | 3.1a — Apply in the same way as main scheme |
| Applies compound angle formula | M1 | 1.1b — Apply in the same way as main scheme |
| $= \left(2\sqrt{3}\cos t\right)^2$ or $12\cos^2 t$ | A1 | 1.1b — Uses correct algebra to find $(x+y)^2 = \left(2\sqrt{3}\cos t\right)^2$ or $(x+y)^2 = 12\cos^2 t$ |
| $(x+y)^2 = 12(1-\sin^2 t) = 12 - 12\sin^2 t = 12 - 12\left(\frac{y}{2}\right)^2$ | M1 | 3.1a — Complete strategy of applying $\cos^2 t + \sin^2 t = 1$ on $(x+y)^2 = \left("2\sqrt{3}\cos t"\right)^2$ to achieve an equation in $x$ and $y$ only |
| $(x+y)^2 + 3y^2 = 12$ | A1 | 2.1 — Correctly proves $(x+y)^2 + ay^2 = b$ with both $a=3$, $b=12$, and no errors seen |
| **(5 marks)** | | |14.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{96e004d9-c6b6-474b-9b67-06e1771c609e-30_659_1232_248_420}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}
Figure 6 shows a sketch of the curve $C$ with parametric equations
$$x = 4 \cos \left( t + \frac { \pi } { 6 } \right) , y = 2 \sin t , \quad 0 < t \leqslant 2 \pi$$
Show that a Cartesian equation of $C$ can be written in the form
$$( x + y ) ^ { 2 } + a y ^ { 2 } = b$$
where $a$ and $b$ are integers to be found.
\hfill \mbox{\textit{Edexcel Paper 1 Q14 [5]}}