| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parametric point verification |
| Difficulty | Moderate -0.8 Part (a) is pure substitution to verify a parametric form (trivial algebra). Part (b) requires recalling the focus of y²=4ax is at (a,0), finding coordinates of A, then calculating a gradient—all routine FP1 techniques with no problem-solving insight needed. Below average difficulty even for Further Maths. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(y^2 = (10r)^2 = 100r^2\) and \(20x = 20 \times 5r^2 = 100r^2\) | B1 | |
| Alternative method: Compare with \(y^2 = 4ax\) and identify \(a = 5\) to give answer. | B1 | (1 mark) |
| (b) Point \(A\) is \((80, 40)\) (stated or seen on diagram). May be given in part (a) | B1, B1 | |
| Focus is \((5, 0)\) (stated or seen on diagram) or \((a, 0)\) with \(a = 5\). May be given in part (a). | ||
| Gradient: \(\frac{40 - 0}{80 - 5} = \frac{40}{75} = \left(\frac{8}{15}\right)\) | M1 A1 | M1: requires use of gradient formula correctly, for their values of \(x\) and \(y\). This mark may be implied by correct answer. A1: Accept \(0.533\) or awrt |
**(a)** $y^2 = (10r)^2 = 100r^2$ and $20x = 20 \times 5r^2 = 100r^2$ | **B1** |
Alternative method: Compare with $y^2 = 4ax$ and identify $a = 5$ to give answer. | **B1** | (1 mark)
**(b)** Point $A$ is $(80, 40)$ (stated or seen on diagram). May be given in part (a) | **B1, B1** |
Focus is $(5, 0)$ (stated or seen on diagram) or $(a, 0)$ with $a = 5$. May be given in part (a). | |
Gradient: $\frac{40 - 0}{80 - 5} = \frac{40}{75} = \left(\frac{8}{15}\right)$ | **M1 A1** | M1: requires use of gradient formula correctly, for their values of $x$ and $y$. This mark may be implied by correct answer. A1: Accept $0.533$ or awrt | (4 marks)
**Total: 5 marks**
**Guidance:**
(a) Allow substitution of $x$ to obtain $y = \pm 10r$ (or just $10r$) or of $y$ to obtain $x$
(b) M1: requires use of gradient formula correctly, for their values of $x$ and $y$. This mark may be implied by correct answer.
Differentiation is **M0 A0**
A1: Accept $0.533$ or awrt
---5. The parabola $C$ has equation $y ^ { 2 } = 20 x$.
\begin{enumerate}[label=(\alph*)]
\item Verify that the point $P \left( 5 t ^ { 2 } , 10 t \right)$ is a general point on $C$.
The point $A$ on $C$ has parameter $t = 4$.\\
The line $l$ passes through $A$ and also passes through the focus of $C$.
\item Find the gradient of $l$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2010 Q5 [5]}}