| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2020 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent/normal meets curve again |
| Difficulty | Standard +0.8 This is a multi-part parametric question requiring finding intersection points, tangent equations, and solving a cubic equation algebraically to find where the tangent meets the curve again. Part (c) involves substituting the tangent equation into parametric equations and solving a cubic, which requires careful algebraic manipulation beyond routine techniques. The question demands multiple connected steps with non-trivial algebra, placing it moderately above average difficulty. |
| 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 |
| \(t^3-4t=0 \Rightarrow t(t^2-4)=0 \Rightarrow t=2\) or \(t=-2\) | M1 | Sets \(t^3-4t=0\) to reach \(t=2\) or \(-2\) |
| \(t=2\): \(x=2\times4-6\times2=-4\), hence \(A=(-4,0)\) | A1 | Substitutes \(t=2\), states coordinates of \(A=(-4,0)\) |
| \(t=-2\): \(x=2\times4-6\times(-2)=20\), \((y=0)\), hence \(B=(20,0)\) | B1* | Show that \(B=(20,0)\). Must show evidence for "20": e.g. \(2\times(-2)^2-6\times(-2)\Rightarrow x=20\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2-4}{4t-6}\) | M1A1 | |
| Sub \(t=-2\): \(\frac{dy}{dx}=\frac{3(4)-4}{4(-2)-6} \Rightarrow\) gradient \(= \left(-\frac{4}{7}\right)\) | M1 | |
| Uses \(\left(-\frac{4}{7}\right)\) and \((20,0)\) to give equation of tangent \(\Rightarrow 7y+4x-80=0\) | M1 A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x=2t^2-6t\), \(y=t^3-4t\) into \(7y+4x-80=0\): \(7(t^3-4t)+4(2t^2-6t)-80=0\) | M1 | |
| \(\Rightarrow 7t^3+8t^2-52t-80=0\) | A1 | |
| \(\Rightarrow (t+2)^2(7t-20)=0\) | ||
| \(t=\frac{20}{7} \Rightarrow x=\ldots\) | dM1 | Dependent on previous M |
| \(x=-\frac{40}{49}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to differentiate \(x(t)\) and \(y(t)\), calculates \(\frac{dy}{dx}\) using \(\frac{dy/dt}{dx/dt}\) | M1 | In part (b) |
| \(\frac{dy}{dx} = \frac{3t^2-4}{4t-6}\) | A1 | |
| Sub \(t = -2\) into \(\frac{dy}{dx}\) to find gradient at \(B\) | M1 | |
| Uses \((20,0)\) and their \(\frac{dy}{dx}\) to find equation of tangent | M1 | If using \(y=mx+c\) must proceed to \(c=\ldots\) |
| \(7y + 4x - 80 = 0\) | A1* | The \(= 0\) must be seen; ISW after sight of this |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(x = 2t^2-6t\), \(y = t^3-4t\) into \(7y+4x-80=0\) | M1 | |
| \(7t^3 + 8t^2 - 52t - 80 = 0\) | A1 | The \(=0\) may be implied by further work |
| Uses correct method to find \(x\) | dM1 | Calculator: accuracy of 2sf for non-exact solutions; \(t=\frac{20}{7}\) is fine. Via factorisation: \(7t^3+8t^2-52t-80=(Pt+a)(Qt+b)(Rt+c)\) with \(PQR=7\) and \(abc=\pm 80\) |
| \(x = -\frac{40}{49}\) | A1 | CSO; no decimals; ignore \(y\) coordinate; penalise extra solutions |
# Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t^3-4t=0 \Rightarrow t(t^2-4)=0 \Rightarrow t=2$ or $t=-2$ | M1 | Sets $t^3-4t=0$ to reach $t=2$ or $-2$ |
| $t=2$: $x=2\times4-6\times2=-4$, hence $A=(-4,0)$ | A1 | Substitutes $t=2$, states coordinates of $A=(-4,0)$ |
| $t=-2$: $x=2\times4-6\times(-2)=20$, $(y=0)$, hence $B=(20,0)$ | B1* | Show that $B=(20,0)$. Must show evidence for "20": e.g. $2\times(-2)^2-6\times(-2)\Rightarrow x=20$ |
---
# Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2-4}{4t-6}$ | M1A1 | |
| Sub $t=-2$: $\frac{dy}{dx}=\frac{3(4)-4}{4(-2)-6} \Rightarrow$ gradient $= \left(-\frac{4}{7}\right)$ | M1 | |
| Uses $\left(-\frac{4}{7}\right)$ and $(20,0)$ to give equation of tangent $\Rightarrow 7y+4x-80=0$ | M1 A1* | |
---
# Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x=2t^2-6t$, $y=t^3-4t$ into $7y+4x-80=0$: $7(t^3-4t)+4(2t^2-6t)-80=0$ | M1 | |
| $\Rightarrow 7t^3+8t^2-52t-80=0$ | A1 | |
| $\Rightarrow (t+2)^2(7t-20)=0$ | | |
| $t=\frac{20}{7} \Rightarrow x=\ldots$ | dM1 | Dependent on previous M |
| $x=-\frac{40}{49}$ | A1 | |
# Question 4 (Parametric Curves):
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to differentiate $x(t)$ and $y(t)$, calculates $\frac{dy}{dx}$ using $\frac{dy/dt}{dx/dt}$ | M1 | In part (b) |
| $\frac{dy}{dx} = \frac{3t^2-4}{4t-6}$ | A1 | |
| Sub $t = -2$ into $\frac{dy}{dx}$ to find gradient at $B$ | M1 | |
| Uses $(20,0)$ and their $\frac{dy}{dx}$ to find equation of tangent | M1 | If using $y=mx+c$ must proceed to $c=\ldots$ |
| $7y + 4x - 80 = 0$ | A1* | The $= 0$ must be seen; ISW after sight of this |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $x = 2t^2-6t$, $y = t^3-4t$ into $7y+4x-80=0$ | M1 | |
| $7t^3 + 8t^2 - 52t - 80 = 0$ | A1 | The $=0$ may be implied by further work |
| Uses correct method to find $x$ | dM1 | Calculator: accuracy of 2sf for non-exact solutions; $t=\frac{20}{7}$ is fine. Via factorisation: $7t^3+8t^2-52t-80=(Pt+a)(Qt+b)(Rt+c)$ with $PQR=7$ and $abc=\pm 80$ |
| $x = -\frac{40}{49}$ | A1 | CSO; no decimals; ignore $y$ coordinate; penalise extra solutions |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{79ac81c3-cd05-4f28-8840-3c8a6960e7b7-10_833_822_127_561}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with parametric equations
$$x = 2 t ^ { 2 } - 6 t , \quad y = t ^ { 3 } - 4 t , \quad t \in \mathbb { R }$$
The curve cuts the $x$-axis at the origin and at the points $A$ and $B$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $A$ and show that $B$ has coordinates (20, 0).
\item Show that the equation of the tangent to the curve at $B$ is
$$7 y + 4 x - 80 = 0$$
The tangent to the curve at $B$ cuts the curve again at the point $P$.
\item Find, using algebra, the $x$ coordinate of $P$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2020 Q4 [12]}}