| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse focus-directrix properties |
| Difficulty | Standard +0.8 This is a Further Maths question requiring knowledge of ellipse foci (using a²-b²=c²) and the focal property that sum of distances equals 2a. Part (a) is straightforward calculation, but part (b) requires understanding the fundamental focus property of ellipses and proving it holds, which goes beyond routine manipulation. The 'show that' proof element and conceptual understanding needed place it moderately above average difficulty. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^24.03d Linear transformations 2D: reflection, rotation, enlargement, shear |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2 = a^2(1-e^2) \Rightarrow 16 = 36(1-e^2) \Rightarrow e = \ldots\) | M1 | Uses \(b^2=a^2(1-e^2)\) with \(a=6\) and \(b=4\) to find \(e^2\) or \(e\) |
| \(e^2 = \frac{20}{36}\) or \(\frac{5}{9}\) or \(e = \frac{\sqrt{5}}{3}\) | A1 | Correct value for \(e\) or \(e^2\) |
| Foci are \(\left(\pm 2\sqrt{5},\ 0\right)\) | A1 | Correct foci |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Perimeter \(= PS + PS' + SS'\) where \(PS+PS' = e(PM+PM') = \ldots\) | M1 | Forms complete strategy using general \(P\); applies focus-directrix property to sides \(PS\) and \(PS'\) |
| \(= e\times\frac{2a}{e} = \ldots\) | M1 | Deduces length of two sides adjacent to \(P\) is constant |
| \(\ldots + 2\times 2\sqrt{5}\) | B1ft | Correct value for \(SS'\) follow through |
| \(= 12 + 4\sqrt{5}\) hence perimeter is constant for any \(P\) on \(E\) | A1* | Correct value with conclusion; dependent on both M marks |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2 = a^2(1-e^2) \Rightarrow 16 = 36(1-e^2) \Rightarrow e = \ldots$ | **M1** | Uses $b^2=a^2(1-e^2)$ with $a=6$ and $b=4$ to find $e^2$ or $e$ |
| $e^2 = \frac{20}{36}$ or $\frac{5}{9}$ or $e = \frac{\sqrt{5}}{3}$ | **A1** | Correct value for $e$ or $e^2$ |
| Foci are $\left(\pm 2\sqrt{5},\ 0\right)$ | **A1** | Correct foci |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Perimeter $= PS + PS' + SS'$ where $PS+PS' = e(PM+PM') = \ldots$ | **M1** | Forms complete strategy using general $P$; applies focus-directrix property to sides $PS$ and $PS'$ |
| $= e\times\frac{2a}{e} = \ldots$ | **M1** | Deduces length of two sides adjacent to $P$ is constant |
| $\ldots + 2\times 2\sqrt{5}$ | **B1ft** | Correct value for $SS'$ follow through |
| $= 12 + 4\sqrt{5}$ hence perimeter is constant for any $P$ on $E$ | **A1*** | Correct value with conclusion; dependent on both M marks |\begin{enumerate}
\item The ellipse $E$ has equation
\end{enumerate}
$$\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 16 } = 1$$
The points $S$ and $S ^ { \prime }$ are the foci of $E$.\\
(a) Find the coordinates of $S$ and $S ^ { \prime }$\\
(b) Show that for any point $P$ on $E$, the triangle $P S S ^ { \prime }$ has constant perimeter and determine its value.
\hfill \mbox{\textit{Edexcel FP1 2020 Q5 [7]}}