| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find tangent equation |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: solving a trigonometric equation for the parameter, then finding dy/dx using the chain rule (dy/dt รท dx/dt) and forming the tangent equation. While it involves multiple steps and careful algebra, all techniques are routine C4 content with no novel problem-solving 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
$$x = \cos 2t, \quad y = \cosec t, \quad 0 < t < \frac{\pi}{2}.$$
The point $P$ on the curve has $x$-coordinate $\frac{1}{2}$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of the parameter $t$ at $P$. [2]
\item Show that the tangent to the curve at $P$ has the equation
$$y = 2x + 1.$$ [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q4 [7]}}