| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find normal equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate implicitly, substitute a point to find the gradient, then find the normal equation. While implicit differentiation is a P4/Further Maths topic (making it slightly harder on an absolute scale), the execution is mechanical with no conceptual challenges or multi-step problem-solving required. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y^2 \rightarrow \ldots y\frac{dy}{dx}\) | M1 | Attempt at chain rule; ... is a constant |
| \(3x^2y \rightarrow \ldots x^2\frac{dy}{dx} + \ldots xy\) | M1 | Attempt at product rule; ... are constants |
| \(2 - 8y\frac{dy}{dx} + 6xy + 3x^2\frac{dy}{dx} = 8x\) | A1 | Correct differentiation o.e. Note: \(\frac{dy}{dx} = 2 - 8y\frac{dy}{dx} + 6xy + 3x^2\frac{dy}{dx} - 8x\) is A0 unless recovered |
| Substitutes \(x=3, y=2 \Rightarrow \frac{dy}{dx} = -\frac{14}{11}\) | M1, A1 | M1: substitutes into suitable equation to find \(\frac{dy}{dx}\); dependent on exactly two \(\frac{dy}{dx}\) terms |
| \(y - 2 = \frac{11}{14}(x-3)\) | dM1 | Correct method for normal using negative reciprocal of their gradient at \((3,2)\) |
| \(11x - 14y - 5 = 0\) | A1 | Accept any integer multiple; "= 0" must be seen |
# Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^2 \rightarrow \ldots y\frac{dy}{dx}$ | M1 | Attempt at chain rule; ... is a constant |
| $3x^2y \rightarrow \ldots x^2\frac{dy}{dx} + \ldots xy$ | M1 | Attempt at product rule; ... are constants |
| $2 - 8y\frac{dy}{dx} + 6xy + 3x^2\frac{dy}{dx} = 8x$ | A1 | Correct differentiation o.e. Note: $\frac{dy}{dx} = 2 - 8y\frac{dy}{dx} + 6xy + 3x^2\frac{dy}{dx} - 8x$ is A0 unless recovered |
| Substitutes $x=3, y=2 \Rightarrow \frac{dy}{dx} = -\frac{14}{11}$ | M1, A1 | M1: substitutes into suitable equation to find $\frac{dy}{dx}$; dependent on exactly two $\frac{dy}{dx}$ terms |
| $y - 2 = \frac{11}{14}(x-3)$ | dM1 | Correct method for normal using negative reciprocal of their gradient at $(3,2)$ |
| $11x - 14y - 5 = 0$ | A1 | Accept any integer multiple; "= 0" must be seen |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$2 x - 4 y ^ { 2 } + 3 x ^ { 2 } y = 4 x ^ { 2 } + 8$$
The point $P ( 3,2 )$ lies on $C$.\\
Find the equation of the normal to $C$ at the point $P$, writing your answer in the form $a x + b y + c = 0$ where $a$, $b$ and $c$ are integers to be found.
\hfill \mbox{\textit{Edexcel P4 2021 Q1 [7]}}