| 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 tangent equation |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring standard techniques: substituting x=0 and y=0 to find axis intersections, then using dy/dx = (dy/dt)/(dx/dt) for the tangent. All steps are routine C4 procedures with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
5.\\
\includegraphics[max width=\textwidth, alt={}, center]{00ad2596-cd76-425d-a373-a0deda11e3c0-2_444_702_246_516}
The diagram shows the curve with parametric equations
$$x = 2 - t ^ { 2 } , \quad y = t ( t + 1 ) , \quad t \geq 0$$
(i) Find the coordinates of the points where the curve meets the coordinate axes.\\
(ii) Find an equation for the tangent to the curve at the point where $t = 2$, giving your answer in the form $a x + b y + c = 0$.\\
\hfill \mbox{\textit{OCR C4 Q5 [9]}}