| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Properties of specific curves |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question requiring routine differentiation (dy/dx = dy/dt รท dx/dt), finding tangent/normal equations, and calculating a triangle area. While multi-step, it follows a predictable template with no novel insights required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x=at^2 \Rightarrow \dot{x}=2at\); \(y=2at \Rightarrow \dot{y}=2a \Rightarrow \frac{dy}{dx}=\frac{1}{t}\); at P grad \(=\frac{1}{p}\) | M1, A1 | |
| Tangent: \((y-2ap)=\frac{1}{p}(x-ap^2)\) | M1 | |
| \(\Rightarrow py-x=ap^2\); when \(y=0,\ x=-ap^2\) | A1 | |
| Normal: \((y-2ap)=-p(x-ap^2)\) | M1 | |
| \(\Rightarrow y+px=ap^3+2ap\); when \(y=0,\ x=ap^2+2a\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \( | TN | =ap^2+2a+ap^2=2a(p^2+1)\) |
| \(\Rightarrow \text{Area}=\frac{1}{2}\times\text{height}\times 2a(p^2+1)=\frac{1}{2}\times 2ap\times 2a(p^2+1)=2a^2p(p^2+1)\) | A1 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x=at^2 \Rightarrow \dot{x}=2at$; $y=2at \Rightarrow \dot{y}=2a \Rightarrow \frac{dy}{dx}=\frac{1}{t}$; at P grad $=\frac{1}{p}$ | M1, A1 | |
| Tangent: $(y-2ap)=\frac{1}{p}(x-ap^2)$ | M1 | |
| $\Rightarrow py-x=ap^2$; when $y=0,\ x=-ap^2$ | A1 | |
| Normal: $(y-2ap)=-p(x-ap^2)$ | M1 | |
| $\Rightarrow y+px=ap^3+2ap$; when $y=0,\ x=ap^2+2a$ | A1 | |
**Total: 6 marks**
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $|TN|=ap^2+2a+ap^2=2a(p^2+1)$ | M1 | |
| $\Rightarrow \text{Area}=\frac{1}{2}\times\text{height}\times 2a(p^2+1)=\frac{1}{2}\times 2ap\times 2a(p^2+1)=2a^2p(p^2+1)$ | A1 | |
**Total: 2 marks**
---5 A curve is given by the parametric equations $x = a t ^ { 2 } , y = 2 a$ (where $a$ is a constant). A point P on the curve has coordinates ( $a p ^ { 2 }$, 2ap).\\
(i) Find the coordinates of the point, T , where the tangent to the curve at P meets the $x$-axis and the coordinates of the point N where the normal to the curve at P meets the $x$-axis.\\
(ii) Hence show that the area of the triangle PTN is $2 a ^ { 2 } p \left( p ^ { 2 } + 1 \right)$ square units.
\hfill \mbox{\textit{OCR MEI C4 Q5 [8]}}