| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2015 |
| Session | June |
| 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 finding intersection points, tangent equations, and solving a cubic equation. While it involves several steps (finding t-values where y=0, computing dy/dx parametrically, finding tangent-curve intersection), each individual technique is standard C3/C4 material. Part (c) requires solving a cubic which factors nicely. The question is more computational than conceptually challenging, placing it moderately above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y=0 \Rightarrow t^3 - 9t = 0 \Rightarrow t(t^2-9)=0 \Rightarrow t=3\) | M1 | Sets \(y=0\) |
| When \(t=3\), \(x=15\) | A1 | Uses \(t=3\) for \(x\) coordinate |
| \(A=(3,0)\) | B1 | States \(A=(3,0)\) with both coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2-9}{2t+2}\) | M1A1 | Differentiates \(x(t)\) and \(y(t)\), calculates \(\frac{dy}{dx}\) |
| Substitutes \(t=3\): gradient \(= \frac{9}{4}\) | M1 | Substitutes \(t=3\) into \(\frac{dy}{dx}\) |
| Uses \(\frac{9}{4}\) and \((15,0)\) to produce \(9x - 4y - 135 = 0\) | M1A1* | Uses point and non-zero gradient for tangent equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \(x=t^2+2t\), \(y=t^3-9t\) into \(9x-4y-135=0\): \(\Rightarrow 4t^3 - 9t^2 - 54t + 135 = 0\) | M1 | Correct substitution to form cubic in \(t\) |
| \((t^2-6t+9)(4t+15)=0\) or \((t-3)^2(4t+15)=0\) | dM1 | Attempts to solve cubic; division by \((t-3)\) or \((t-3)^2\) |
| \(t = -\frac{15}{4}\) | A1 | |
| Coordinates of \(X\) are \(\left(\frac{105}{16}, -\frac{1215}{64}\right)\) | ddM1A1 cso | Uses \(t\) value for both coordinates; dependent on both M marks |
# Question 9:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y=0 \Rightarrow t^3 - 9t = 0 \Rightarrow t(t^2-9)=0 \Rightarrow t=3$ | M1 | Sets $y=0$ |
| When $t=3$, $x=15$ | A1 | Uses $t=3$ for $x$ coordinate |
| $A=(3,0)$ | B1 | States $A=(3,0)$ with both coordinates |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2-9}{2t+2}$ | M1A1 | Differentiates $x(t)$ and $y(t)$, calculates $\frac{dy}{dx}$ |
| Substitutes $t=3$: gradient $= \frac{9}{4}$ | M1 | Substitutes $t=3$ into $\frac{dy}{dx}$ |
| Uses $\frac{9}{4}$ and $(15,0)$ to produce $9x - 4y - 135 = 0$ | M1A1* | Uses point and non-zero gradient for tangent equation |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $x=t^2+2t$, $y=t^3-9t$ into $9x-4y-135=0$: $\Rightarrow 4t^3 - 9t^2 - 54t + 135 = 0$ | M1 | Correct substitution to form cubic in $t$ |
| $(t^2-6t+9)(4t+15)=0$ or $(t-3)^2(4t+15)=0$ | dM1 | Attempts to solve cubic; division by $(t-3)$ or $(t-3)^2$ |
| $t = -\frac{15}{4}$ | A1 | |
| Coordinates of $X$ are $\left(\frac{105}{16}, -\frac{1215}{64}\right)$ | ddM1A1 cso | Uses $t$ value for both coordinates; dependent on both M marks |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c08fbab-283e-4c92-89a4-10f68f37e133-14_709_824_118_559}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with parametric equations
$$x = t ^ { 2 } + 2 t , \quad y = t ^ { 3 } - 9 t , \quad t \in \mathbb { R }$$
The curve cuts the $x$-axis at the origin and at the points $A$ and $B$ as shown in Figure 3 .
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of point $A$ and show that point $B$ has coordinates ( 15,0 ).
\item Show that the equation of the tangent to the curve at $B$ is $9 x - 4 y - 135 = 0$
The tangent to the curve at $B$ cuts the curve again at the point $X$.
\item Find the coordinates of $X$.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2015 Q9 [13]}}