| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 9 |
| 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 differentiation application requiring chain rule, finding tangent/normal equations using standard formulas, and basic algebraic manipulation. Part (a) is a 'show that' which guides students, and part (b) follows directly from standard normal line procedures. Slightly easier than average due to the scaffolding and routine nature of the techniques. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(x = 3, \quad y = \sqrt{20} = 2\sqrt{5}\) | B1 |
| \(\frac{dy}{dx} = \frac{1}{3}(3x+11)^{-\frac{1}{3}} \times 3 = \frac{1}{3}(3x+11)^{-\frac{1}{3}}\) | M1 A1 | |
| \(\text{grad} = \frac{3}{4\sqrt{5}}\) | A1 | |
| \(y - 2\sqrt{5} = \frac{3}{4\sqrt{5}}(x-3)\) | M1 | |
| \(4\sqrt{5}y - 40 = 3x - 9\) | ||
| \(3x - 4\sqrt{5}y + 31 = 0\) | A1 | |
| (b) | Normal: \(y - 2\sqrt{5} = -\frac{4\sqrt{5}}{3}(x-3)\) | M1 |
| At \(Q\), \(x = 0\): \(y - 2\sqrt{5} = 4\sqrt{5}, \quad y = 6\sqrt{5}\) | M1 A1 | |
| (9) |
(a) | $x = 3, \quad y = \sqrt{20} = 2\sqrt{5}$ | B1 |
| $\frac{dy}{dx} = \frac{1}{3}(3x+11)^{-\frac{1}{3}} \times 3 = \frac{1}{3}(3x+11)^{-\frac{1}{3}}$ | M1 A1 |
| $\text{grad} = \frac{3}{4\sqrt{5}}$ | A1 |
| $y - 2\sqrt{5} = \frac{3}{4\sqrt{5}}(x-3)$ | M1 |
| $4\sqrt{5}y - 40 = 3x - 9$ | |
| $3x - 4\sqrt{5}y + 31 = 0$ | A1 |
(b) | Normal: $y - 2\sqrt{5} = -\frac{4\sqrt{5}}{3}(x-3)$ | M1 |
| At $Q$, $x = 0$: $y - 2\sqrt{5} = 4\sqrt{5}, \quad y = 6\sqrt{5}$ | M1 A1 |
| | (9) |2. A curve has the equation $y = \sqrt { 3 x + 11 }$.
The point $P$ on the curve has $x$-coordinate 3 .
\begin{enumerate}[label=(\alph*)]
\item Show that the tangent to the curve at $P$ has the equation
$$3 x - 4 \sqrt { 5 } y + 31 = 0$$
The normal to the curve at $P$ crosses the $y$-axis at $Q$.
\item Find the $y$-coordinate of $Q$ in the form $k \sqrt { 5 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q2 [9]}}