| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| 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 differentiation application requiring chain rule, finding gradient of normal (negative reciprocal), and substituting x=0. All steps are routine C3 techniques with no conceptual challenges, making it slightly easier than average but not trivial due to the surd manipulation required. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
5. A curve has the equation $y = \sqrt { 3 x + 11 }$.
The point $P$ on the curve has $x$-coordinate 3 .\\
(i) 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 $Q$.\\
(ii) Find the $y$-coordinate of $Q$ in the form $k \sqrt { 5 }$.\\
\hfill \mbox{\textit{OCR C3 Q5 [8]}}