| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find vertical tangent points |
| Difficulty | Standard +0.8 This question requires implicit differentiation (standard C4 technique) but part (b) demands conceptual understanding that vertical tangents occur when dx/dy = 0 (or dy/dx is undefined), then solving a simultaneous system with the original curve equation. The insight that 'parallel to y-axis' means the denominator of dy/dx equals zero is non-trivial for many students, elevating this above routine implicit differentiation exercises. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x + \left(4y + 4x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0\) | M1 A1 B1 | M1: implicit differentiation including \(\pm ky\frac{dy}{dx}\); A1: \(x^2 \to 2x\) and \(2y\frac{dy}{dx}\) correct; B1: \(4y + 4x\frac{dy}{dx}\) correct |
| \(2x + 4y + (4x+2y)\frac{dy}{dx} = 0\) | dM1 | Factorises out \(\frac{dy}{dx}\) (at least two \(\frac{dy}{dx}\) terms present) |
| \(\frac{dy}{dx} = \frac{-2x-4y}{4x+2y}\left\{=\frac{-x-2y}{2x+y}\right\}\) | A1 cso oe | Fully correct simplified expression; cso — not awarded if solution not completely correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4x + 2y = 0 \Rightarrow y = -2x\) or \(x = -\frac{1}{2}y\) | M1, A1 | Sets numerator of \(\frac{dy}{dx}\) = 0 ... wait, sets denominator; M1 for \(4x+2y=0\) |
| Substitutes into \(x^2+4xy+y^2+27=0\): e.g. \(x^2+4x(-2x)+(-2x)^2+27=0 \Rightarrow -3x^2+27=0\) | M1* | Substitutes into original curve equation |
| \(x^2 = 9 \Rightarrow x = -3\) (or \(y^2=36 \Rightarrow y=6\)) | dM1* | Solves to find \(x\) or \(y\) value |
| \(x = -3\) | A1 | — |
| When \(x=-3\): \(y = -2(-3) = 6\); coordinates \((-3, 6)\) | ddM1*, A1 cso | Finds both coordinates; dependent on all previous M marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets denominator of \(\frac{dy}{dx}\) equal to zero (or numerator of \(\frac{dx}{dy}\) equal to zero) | M1 | oe |
| Rearranges to give either \(y = -2x\) or \(x = -\frac{1}{2}y\) | A1 | Correct solution only |
| Substitutes \(y = \pm\lambda x\) or \(x = \pm\mu y\) or \(y = \pm\lambda x \pm a\) or \(x = \pm\mu y \pm b\) \((\lambda \neq 0, \mu \neq 0)\) into \(x^2 + 4xy + y^2 + 27 = 0\) | M1* | Forms equation in one variable |
| Leading to at least either \(x^2 = A,\ A>0\) or \(y^2 = B,\ B>0\) | dM1* | Dependent on M1* |
| For \(x = -3\) (ignore \(x=3\)) or if \(y\) found first, \(y = 6\) (ignore \(y = -6\)) | A1 | Correct solution only |
| Substitutes value of \(x\) into \(y = \pm\lambda x\) to give \(y\) value | ddM1* | Dependent on M1* and dM1* |
| \((-3,\ 6)\) | A1 | cso |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x + \left(4y + 4x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0$ | M1 A1 B1 | M1: implicit differentiation including $\pm ky\frac{dy}{dx}$; A1: $x^2 \to 2x$ and $2y\frac{dy}{dx}$ correct; B1: $4y + 4x\frac{dy}{dx}$ correct |
| $2x + 4y + (4x+2y)\frac{dy}{dx} = 0$ | dM1 | Factorises out $\frac{dy}{dx}$ (at least two $\frac{dy}{dx}$ terms present) |
| $\frac{dy}{dx} = \frac{-2x-4y}{4x+2y}\left\{=\frac{-x-2y}{2x+y}\right\}$ | A1 cso oe | Fully correct simplified expression; cso — not awarded if solution not completely correct |
**[5 marks]**
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4x + 2y = 0 \Rightarrow y = -2x$ or $x = -\frac{1}{2}y$ | M1, A1 | Sets numerator of $\frac{dy}{dx}$ = 0 ... wait, sets denominator; M1 for $4x+2y=0$ |
| Substitutes into $x^2+4xy+y^2+27=0$: e.g. $x^2+4x(-2x)+(-2x)^2+27=0 \Rightarrow -3x^2+27=0$ | M1* | Substitutes into original curve equation |
| $x^2 = 9 \Rightarrow x = -3$ (or $y^2=36 \Rightarrow y=6$) | dM1* | Solves to find $x$ or $y$ value |
| $x = -3$ | A1 | — |
| When $x=-3$: $y = -2(-3) = 6$; coordinates $(-3, 6)$ | ddM1*, A1 cso | Finds both coordinates; dependent on all previous M marks |
**[7 marks]**
# Question 7 (Part b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets denominator of $\frac{dy}{dx}$ equal to zero (or numerator of $\frac{dx}{dy}$ equal to zero) | M1 | oe |
| Rearranges to give either $y = -2x$ or $x = -\frac{1}{2}y$ | A1 | Correct solution only |
| Substitutes $y = \pm\lambda x$ or $x = \pm\mu y$ or $y = \pm\lambda x \pm a$ or $x = \pm\mu y \pm b$ $(\lambda \neq 0, \mu \neq 0)$ into $x^2 + 4xy + y^2 + 27 = 0$ | M1* | Forms equation in one variable |
| Leading to at least either $x^2 = A,\ A>0$ or $y^2 = B,\ B>0$ | dM1* | Dependent on M1* |
| For $x = -3$ (ignore $x=3$) or if $y$ found first, $y = 6$ (ignore $y = -6$) | A1 | Correct solution only |
| Substitutes value of $x$ into $y = \pm\lambda x$ to give $y$ value | ddM1* | Dependent on M1* and dM1* |
| $(-3,\ 6)$ | A1 | cso |
---7. A curve is described by the equation
$$x ^ { 2 } + 4 x y + y ^ { 2 } + 27 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
A point $Q$ lies on the curve.\\
The tangent to the curve at $Q$ is parallel to the $y$-axis.\\
Given that the $x$ coordinate of $Q$ is negative,
\item use your answer to part (a) to find the coordinates of $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2013 Q7 [12]}}