| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate term-by-term (including the product rule for 6xy), substitute the point to find dy/dx, then write the tangent equation. While it involves multiple techniques, it follows a standard algorithmic procedure with no conceptual surprises, making it slightly easier than average for C4 level. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Differentiates to obtain \(6x + 8y\frac{dy}{dx} - 2\) | M1, A1 | |
| \(+(6x\frac{dy}{dx} + 6y) = 0\) | +(B1) | |
| \(\frac{dy}{dx} = \frac{2-6x-6y}{6x+8y}\) | ||
| Substitutes \(x=1\), \(y=-2\) into expression involving \(\frac{dy}{dx}\) to give \(\frac{dy}{dx} = -\frac{8}{10}\) | M1, A1 | |
| Uses line equation with numerical gradient: \(y-(-2) = (\text{their gradient})(x-1)\) or finds \(c\) and uses \(y = (\text{their gradient})x + c\) | M1 | |
| \(5y + 4x + 6 = 0\) (or equivalent \(= 0\)) | A1\(\sqrt{}\) | [7] |
## Question 1:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Differentiates to obtain $6x + 8y\frac{dy}{dx} - 2$ | M1, A1 | |
| $+(6x\frac{dy}{dx} + 6y) = 0$ | +(B1) | |
| $\frac{dy}{dx} = \frac{2-6x-6y}{6x+8y}$ | | |
| Substitutes $x=1$, $y=-2$ into expression involving $\frac{dy}{dx}$ to give $\frac{dy}{dx} = -\frac{8}{10}$ | M1, A1 | |
| Uses line equation with numerical gradient: $y-(-2) = (\text{their gradient})(x-1)$ or finds $c$ and uses $y = (\text{their gradient})x + c$ | M1 | |
| $5y + 4x + 6 = 0$ (or equivalent $= 0$) | A1$\sqrt{}$ | **[7]** |
---\begin{enumerate}
\item A curve $C$ is described by the equation
\end{enumerate}
$$3 x ^ { 2 } + 4 y ^ { 2 } - 2 x + 6 x y - 5 = 0$$
Find an equation of the tangent to $C$ at the point $( 1 , - 2 )$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
\hfill \mbox{\textit{Edexcel C4 2006 Q1 [7]}}