| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Normal meets curve/axis — further geometry |
| Difficulty | Moderate -0.3 This is a straightforward C1 question requiring basic differentiation to find the gradient, then using perpendicular gradient properties for the normal, followed by solving a quadratic equation. While it involves multiple steps (8 marks total), each step uses standard techniques with no conceptual challenges beyond routine application of tangent/normal theory. |
| Spec | 1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = 2x + 2\) | M1 A1 | |
| grad of tangent \(= 2\) | A1 | |
| grad of normal \(= \frac{-1}{2} = -\frac{1}{2}\) | M1 | |
| \(\therefore y = -\frac{1}{4}x\) | A1 | |
| (b) \(x^2 + 2x = -\frac{1}{4}x\) | ||
| \(2x^2 + 5x = 0, \quad x(2x + 5) = 0\) | M1 | |
| \(x = 0\) (at O), \(-\frac{5}{2}\) | A1 | |
| \(\therefore \left(-\frac{5}{2}, \frac{5}{4}\right)\) | A1 | (8 marks) |
**(a)** $\frac{dy}{dx} = 2x + 2$ | M1 A1 |
grad of tangent $= 2$ | A1 |
grad of normal $= \frac{-1}{2} = -\frac{1}{2}$ | M1 |
$\therefore y = -\frac{1}{4}x$ | A1 |
**(b)** $x^2 + 2x = -\frac{1}{4}x$ | |
$2x^2 + 5x = 0, \quad x(2x + 5) = 0$ | M1 |
$x = 0$ (at O), $-\frac{5}{2}$ | A1 |
$\therefore \left(-\frac{5}{2}, \frac{5}{4}\right)$ | A1 | (8 marks)The curve with equation $y = x^2 + 2x$ passes through the origin, $O$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the normal to the curve at $O$. [5]
\item Find the coordinates of the point where the normal to the curve at $O$ intersects the curve again. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q6 [8]}}