| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent/normal meets curve again |
| Difficulty | Challenging +1.2 This is a multi-part parametric question requiring standard differentiation (dy/dx = dy/dt รท dx/dt), finding a normal equation, then solving a simultaneous equation with parametric forms. Part (c) requires substituting the linear equation into parametric equations and solving a trigonometric equation, which is moderately challenging but follows standard A-level techniques. The 'show that' in part (b) provides scaffolding, and while part (c) requires careful algebraic manipulation, it doesn't demand novel insight beyond applying learned methods systematically. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) | M1 | Achieves a form \(k\frac{\sin 2t}{\sin t}\) or \(k\frac{\sin t \cos t}{\sin t}\) |
| \(\frac{dy}{dx} = \frac{\sqrt{3}\sin 2t}{\sin t}\) \((= 2\sqrt{3}\cos t)\) | A1 | Either form accepted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(t = \frac{2\pi}{3}\) in \(\frac{dy}{dx} = \frac{\sqrt{3}\sin 2t}{\sin t} = (-\sqrt{3})\) | M1 | Must substitute into expression in terms of \(t\) |
| Uses gradient of normal \(= -\frac{1}{dy/dx} = \left(\frac{1}{\sqrt{3}}\right)\) | M1 | Negative reciprocal; may be seen in equation of line |
| Coordinates of \(P = \left(-1, -\frac{\sqrt{3}}{2}\right)\) | B1 | States or uses in tangent or normal |
| Correct form of normal: \(y + \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}(x+1)\) | M1 | Uses numerical value of \(-1/(dy/dx)\) with \(P\) to form equation |
| Completes proof \(\Rightarrow 2x - 2\sqrt{3}y - 1 = 0\) | A1* | All aspects correct; correct answer only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x = 2\cos t\) and \(y = \sqrt{3}\cos 2t\) into \(2x - 2\sqrt{3}y - 1 = 0\) | M1 | Alternatively use \(\cos 2t = 2\cos^2 t -1\) to set up \(y = Ax^2 + B\) |
| Uses identity \(\cos 2t = 2\cos^2 t - 1\) to produce quadratic in \(\cos t\) | M1 | In alternative: combine \(y = Ax^2+B\) with \(2x-2\sqrt{3}y-1=0\) |
| \(\Rightarrow 12\cos^2 t - 4\cos t - 5 = 0\) | A1 | Alternatively \(3x^2 - 2x - 5 = 0\) or \(12\sqrt{3}y^2 + 4y - 7\sqrt{3} = 0\) |
| Finds \(\cos t = \frac{5}{6}\), \(\not\frac{1}{2}\) (reject) | M1 | Solves quadratic and rejects value corresponding to \(P\) |
| Substitutes \(\cos t = \frac{5}{6}\) into \(x = 2\cos t\), \(y = \sqrt{3}\cos 2t\) | M1 | For finding the other coordinate once one is known |
| \(Q = \left(\frac{5}{3}, \frac{7}{18}\sqrt{3}\right)\) | A1 | Allow \(x = \frac{5}{3}\), \(y = \frac{7}{18}\sqrt{3}\); not decimal equivalents |
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | M1 | Achieves a form $k\frac{\sin 2t}{\sin t}$ or $k\frac{\sin t \cos t}{\sin t}$ |
| $\frac{dy}{dx} = \frac{\sqrt{3}\sin 2t}{\sin t}$ $(= 2\sqrt{3}\cos t)$ | A1 | Either form accepted |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $t = \frac{2\pi}{3}$ in $\frac{dy}{dx} = \frac{\sqrt{3}\sin 2t}{\sin t} = (-\sqrt{3})$ | M1 | Must substitute into expression in terms of $t$ |
| Uses gradient of normal $= -\frac{1}{dy/dx} = \left(\frac{1}{\sqrt{3}}\right)$ | M1 | Negative reciprocal; may be seen in equation of line |
| Coordinates of $P = \left(-1, -\frac{\sqrt{3}}{2}\right)$ | B1 | States or uses in tangent or normal |
| Correct form of normal: $y + \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}}(x+1)$ | M1 | Uses numerical value of $-1/(dy/dx)$ with $P$ to form equation |
| Completes proof $\Rightarrow 2x - 2\sqrt{3}y - 1 = 0$ | A1* | All aspects correct; correct answer only |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x = 2\cos t$ and $y = \sqrt{3}\cos 2t$ into $2x - 2\sqrt{3}y - 1 = 0$ | M1 | Alternatively use $\cos 2t = 2\cos^2 t -1$ to set up $y = Ax^2 + B$ |
| Uses identity $\cos 2t = 2\cos^2 t - 1$ to produce quadratic in $\cos t$ | M1 | In alternative: combine $y = Ax^2+B$ with $2x-2\sqrt{3}y-1=0$ |
| $\Rightarrow 12\cos^2 t - 4\cos t - 5 = 0$ | A1 | Alternatively $3x^2 - 2x - 5 = 0$ or $12\sqrt{3}y^2 + 4y - 7\sqrt{3} = 0$ |
| Finds $\cos t = \frac{5}{6}$, $\not\frac{1}{2}$ (reject) | M1 | Solves quadratic and rejects value corresponding to $P$ |
| Substitutes $\cos t = \frac{5}{6}$ into $x = 2\cos t$, $y = \sqrt{3}\cos 2t$ | M1 | For finding the other coordinate once one is known |
| $Q = \left(\frac{5}{3}, \frac{7}{18}\sqrt{3}\right)$ | A1 | Allow $x = \frac{5}{3}$, $y = \frac{7}{18}\sqrt{3}$; not decimal equivalents |
---\begin{enumerate}
\item The curve $C$ has parametric equations
\end{enumerate}
$$x = 2 \cos t , \quad y = \sqrt { 3 } \cos 2 t , \quad 0 \leqslant t \leqslant \pi$$
(a) Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.
The point $P$ lies on $C$ where $t = \frac { 2 \pi } { 3 }$\\
The line $l$ is the normal to $C$ at $P$.\\
(b) Show that an equation for $l$ is
$$2 x - 2 \sqrt { 3 } y - 1 = 0$$
The line $l$ intersects the curve $C$ again at the point $Q$.\\
(c) Find the exact coordinates of $Q$.
You must show clearly how you obtained your answers.
\hfill \mbox{\textit{Edexcel Paper 1 Q13 [13]}}