| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 10 |
| 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 standard implicit differentiation question requiring students to find dy/dx, calculate the gradient at a point, find the normal equation, then solve simultaneously with the original curve equation. While it involves multiple steps (5-6 marks typical), each step follows routine procedures taught in C4 with no novel insight required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6x - 2 + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\) | M1 A1 | |
| \((-1, 3) \Rightarrow -6 - 2 + 3 - \frac{dy}{dx} + 6\frac{dy}{dx} = 0\), \(\frac{dy}{dx} = 1\) | M1 A1 | |
| grad of normal \(= -1\), \(\therefore y - 3 = -(x + 1)\) | M1 | |
| \(y = 2 - x\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| sub. \(\Rightarrow 3x^2 - 2x + x(2-x) + (2-x)^2 - 11 = 0\) | M1 | |
| \(3x^2 - 4x - 7 = 0\) | A1 | |
| \((3x - 7)(x + 1) = 0\) | M1 | |
| \(x = -1\) (at \(P\)) or \(\frac{7}{3}\) \(\therefore \left(\frac{7}{3}, -\frac{1}{3}\right)\) | A1 | (10) |
# Question 7:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6x - 2 + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$ | M1 A1 | |
| $(-1, 3) \Rightarrow -6 - 2 + 3 - \frac{dy}{dx} + 6\frac{dy}{dx} = 0$, $\frac{dy}{dx} = 1$ | M1 A1 | |
| grad of normal $= -1$, $\therefore y - 3 = -(x + 1)$ | M1 | |
| $y = 2 - x$ | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| sub. $\Rightarrow 3x^2 - 2x + x(2-x) + (2-x)^2 - 11 = 0$ | M1 | |
| $3x^2 - 4x - 7 = 0$ | A1 | |
| $(3x - 7)(x + 1) = 0$ | M1 | |
| $x = -1$ (at $P$) or $\frac{7}{3}$ $\therefore \left(\frac{7}{3}, -\frac{1}{3}\right)$ | A1 | **(10)** |
---\begin{enumerate}
\item A curve has the equation
\end{enumerate}
$$3 x ^ { 2 } - 2 x + x y + y ^ { 2 } - 11 = 0$$
The point $P$ on the curve has coordinates $( - 1,3 )$.\\
(i) Show that the normal to the curve at $P$ has the equation $y = 2 - x$.\\
(ii) Find the coordinates of the point where the normal to the curve at $P$ meets the curve again.\\
\hfill \mbox{\textit{OCR C4 Q7 [10]}}