| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find normal equation |
| Difficulty | Standard +0.3 This is a standard C4 parametric differentiation question requiring routine application of dy/dx = (dy/dt)/(dx/dt), finding the normal gradient, and eliminating the parameter. While it involves multiple steps (6+3 marks), all techniques are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
A curve has parametric equations
$$x = t^3 + 1, \quad y = \frac{2}{t}, \quad t \neq 0.$$
\begin{enumerate}[label=(\roman*)]
\item Find an equation for the normal to the curve at the point where $t = 1$, giving your answer in the form $y = mx + c$. [6]
\item Find a cartesian equation for the curve in the form $y = f(x)$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q4 [9]}}