| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find normal equation |
| Difficulty | Moderate -0.3 This is a straightforward parametric equations question requiring standard techniques: chain rule for dy/dx, finding a normal line equation, and eliminating the parameter (here simply t = y/4, substitute into x equation). All three parts are routine C4 exercises with no conceptual challenges, 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to use \(\frac{dv}{dt}\) or \(\frac{dy}{dr} \cdot \frac{dr}{dx}\) | M1 | Not just quote formula |
| \(\frac{4}{2t}\) or \(\frac{2}{t}\) | A1 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Subst \(t = 4\) into their (i), invert & change sign | M1 | |
| Subst \(t = 4\) into \((x,y)\) & use num grad for tgt/normal | M1 | |
| \(y = -2x + 52\) AEF CAO (no f.t.) | A1 3 | Only the eqn of normal accepted |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to eliminate \(t\) from the 2 given equations | M1 | |
| \(x = 2 + \frac{y^2}{16}\) or \(y^2 = 16(x-2)\) AEF ISW | A1 2 | Mark at earliest acceptable form. |
**(i)**
Attempt to use $\frac{dv}{dt}$ or $\frac{dy}{dr} \cdot \frac{dr}{dx}$ | M1 | Not just quote formula
$\frac{4}{2t}$ or $\frac{2}{t}$ | A1 2 |
**(ii)**
Subst $t = 4$ into their (i), invert & change sign | M1 |
Subst $t = 4$ into $(x,y)$ & use num grad for tgt/normal | M1 |
$y = -2x + 52$ AEF CAO (no f.t.) | A1 3 | Only the eqn of normal accepted
**(iii)**
Attempt to eliminate $t$ from the 2 given equations | M1 |
$x = 2 + \frac{y^2}{16}$ or $y^2 = 16(x-2)$ AEF ISW | A1 2 | Mark at earliest acceptable form.
---4 A curve has parametric equations
$$x = 2 + t ^ { 2 } , \quad y = 4 t$$
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.\\
(ii) Find the equation of the normal at the point where $t = 4$, giving your answer in the form $y = m x + c$.\\
(iii) Find a cartesian equation of the curve.
\hfill \mbox{\textit{OCR C4 2011 Q4 [7]}}