| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parametric point verification |
| Difficulty | Moderate -0.8 This is a straightforward Further Maths question requiring basic parametric-to-Cartesian conversion (eliminating t to get xy=25) and simple coordinate substitution to find two points then their midpoint. While it's from FP1, the techniques are routine with no problem-solving or insight required—just mechanical application of standard methods. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(xy = 25 = 5^2\) or \(c = \pm 5\) | B1 | \(xy=25\) only B1; \(c^2=25\) only B1; \(c=5\) only B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A\) has co-ords \((5,5)\) and \(B\) has co-ords \((25,1)\) | B1 | Both coordinates required for B1 |
| Mid point is at \((15, 3)\) | M1 A1 | Add coordinates and divide by 2 on both for M1 |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $xy = 25 = 5^2$ or $c = \pm 5$ | B1 | $xy=25$ only B1; $c^2=25$ only B1; $c=5$ only B1 |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A$ has co-ords $(5,5)$ and $B$ has co-ords $(25,1)$ | B1 | Both coordinates required for B1 |
| Mid point is at $(15, 3)$ | M1 A1 | Add coordinates and divide by 2 on both for M1 |
---3. The rectangular hyperbola, $H$, has parametric equations $x = 5 t , y = \frac { 5 } { t } , t \neq 0$.
\begin{enumerate}[label=(\alph*)]
\item Write the cartesian equation of $H$ in the form $x y = c ^ { 2 }$.
Points $A$ and $B$ on the hyperbola have parameters $t = 1$ and $t = 5$ respectively.
\item Find the coordinates of the mid-point of $A B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2009 Q3 [4]}}