| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse focus-directrix properties |
| Difficulty | Challenging +1.2 This is a structured multi-part question on ellipse focus-directrix properties requiring knowledge of eccentricity relationships and the focal-directrix definition. While it involves Further Maths content (making it inherently harder than Core), the question is highly scaffolded with clear steps leading to the proof in part (c). Parts (a)-(b) are standard recall/application, part (c) is a guided algebraic proof, and parts (d)-(e) are routine calculations. The scaffolding and standard nature of the techniques place it slightly above average difficulty overall. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| \((\pm 7e,\ 0)\) | B1 | Correct coordinates or \(x = \pm 7e,\ y = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \pm\frac{7}{e}\) | B1 | Correct equations. SC: If "49" used for "7" consistently in (i) and (ii) score B0 B1. |
| Answer | Marks | Guidance |
|---|---|---|
| \(PS^2 = (x - {'}7e{'})^2 + y^2\) | B1ft | Correct expression or equivalent with their \(7e\). Must be in terms of \(e\), \(x\) and \(y\) only. Apply isw once correct expression seen. |
| Answer | Marks | Guidance |
|---|---|---|
| \(PM^2 = \left(\frac{7}{e} - x\right)^2\) | B1ft | Correct expression or equivalent with their \(\frac{7}{e}\). Must be in terms of \(e\) and \(x\) only. Apply isw once correct expression seen. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow x^2 - 14ex + 49e^2 + y^2 = 49 - 14ex + e^2x^2\) | M1 | Applies \(PS^2 = e^2 PM^2\) with their \(PS\) and \(PM\) and expands (condone poor squaring). |
| \(x^2(1-e^2) + y^2 = 49(1-e^2) \Rightarrow \frac{x^2}{49} + \frac{y^2}{49(1-e^2)} = 1 \Rightarrow b^2 = 49(1-e^2)\)* | A1* | Reaches given answer with fully correct proof. All shown steps required. Note possible to obtain this result even if B marks not scored in (b). |
| Answer | Marks | Guidance |
|---|---|---|
| \((4\sqrt{3})^2 = 49(1-e^2) \Rightarrow e^2\ldots\) or \(e = \ldots\) | M1 | Replaces \(b^2\) with \((4\sqrt{3})^2\) and solves for \(e^2\) or \(e\). |
| \(e = \frac{1}{7}\) | A1 | Correct exact value for \(e\) (not \(\pm\)). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \frac{7}{2} \Rightarrow \frac{(\frac{7}{2})^2}{49} + \frac{y^2}{48} = 1 \Rightarrow y = \pm 6\) | M1 | Substitutes \(x = \frac{7}{2}\) into ellipse equation and obtains a value for \(y\) |
| Area \(\Delta OPM = \left(\frac{1}{2}\right)\left(\frac{7}{(\frac{1}{e})}, -\frac{7}{2}\right)(6) = \ldots\) | dM1 | Correct method for area of triangle \(OPM\) with their \(\frac{7}{e}\) and their 6; may use shoelace method e.g. \(\frac{1}{2}\begin{vmatrix}3.5 & 0 & 49 & 3.5\\6 & 0 & 6 & 6\end{vmatrix} = \frac{1}{2}(49\times6 - 6\times3.5)\) |
| \(\frac{273}{2}\) or \(136\frac{1}{2}\) or \(136.5\) | A1 | Any correct exact value |
| Special Case: \(x=\frac{7}{2} \Rightarrow y=36 \Rightarrow\) Area \(= \ldots(819)\) scores M0M1A0 | — | — |
## Question 3(a)(i):
$(\pm 7e,\ 0)$ | B1 | Correct **coordinates** or $x = \pm 7e,\ y = 0$
## Question 3(a)(ii):
$x = \pm\frac{7}{e}$ | B1 | Correct **equations**. SC: If "49" used for "7" consistently in (i) and (ii) score B0 B1.
## Question 3(b)(i):
$PS^2 = (x - {'}7e{'})^2 + y^2$ | B1ft | Correct expression or equivalent with their $7e$. Must be in terms of $e$, $x$ and $y$ only. Apply isw once correct expression seen.
## Question 3(b)(ii):
$PM^2 = \left(\frac{7}{e} - x\right)^2$ | B1ft | Correct expression or equivalent with their $\frac{7}{e}$. Must be in terms of $e$ and $x$ only. Apply isw once correct expression seen.
## Question 3(c):
$\frac{PS}{PM} = e \Rightarrow PS^2 = e^2 PM^2 \Rightarrow (x-{'}7e{'})^2 + y^2 = e^2\left(\frac{7}{e}-x\right)^2$
$\Rightarrow x^2 - 14ex + 49e^2 + y^2 = 49 - 14ex + e^2x^2$ | M1 | Applies $PS^2 = e^2 PM^2$ with their $PS$ and $PM$ and expands (condone poor squaring).
$x^2(1-e^2) + y^2 = 49(1-e^2) \Rightarrow \frac{x^2}{49} + \frac{y^2}{49(1-e^2)} = 1 \Rightarrow b^2 = 49(1-e^2)$* | A1* | Reaches given answer with fully correct proof. All shown steps required. Note possible to obtain this result even if B marks not scored in (b).
## Question 3(d):
$(4\sqrt{3})^2 = 49(1-e^2) \Rightarrow e^2\ldots$ or $e = \ldots$ | M1 | Replaces $b^2$ with $(4\sqrt{3})^2$ and solves for $e^2$ or $e$.
$e = \frac{1}{7}$ | A1 | Correct exact value for $e$ (not $\pm$).
## Question (e) [Ellipse Area]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \frac{7}{2} \Rightarrow \frac{(\frac{7}{2})^2}{49} + \frac{y^2}{48} = 1 \Rightarrow y = \pm 6$ | M1 | Substitutes $x = \frac{7}{2}$ into ellipse equation and obtains a value for $y$ |
| Area $\Delta OPM = \left(\frac{1}{2}\right)\left(\frac{7}{(\frac{1}{e})}, -\frac{7}{2}\right)(6) = \ldots$ | dM1 | Correct method for area of triangle $OPM$ with their $\frac{7}{e}$ and their 6; may use shoelace method e.g. $\frac{1}{2}\begin{vmatrix}3.5 & 0 & 49 & 3.5\\6 & 0 & 6 & 6\end{vmatrix} = \frac{1}{2}(49\times6 - 6\times3.5)$ |
| $\frac{273}{2}$ or $136\frac{1}{2}$ or $136.5$ | A1 | Any correct exact value |
| **Special Case:** $x=\frac{7}{2} \Rightarrow y=36 \Rightarrow$ Area $= \ldots(819)$ scores M0M1A0 | — | — |
---\begin{enumerate}
\item The ellipse $E$ has equation
\end{enumerate}
$$\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$$
where $b$ is a constant and $0 < b < 7$\\
The eccentricity of the ellipse is $e$\\
(a) Write down, in terms of $e$ only,\\
(i) the coordinates of the foci of $E$\\
(ii) the equations of the directrices of $E$
Given that
\begin{itemize}
\item the point $P ( x , y )$ lies on $E$ where $x > 0$
\item the point $S$ is the focus of $E$ on the positive $x$-axis
\item the line $l$ is the directrix of $E$ which crosses the positive $x$-axis
\item the point $M$ lies on $l$ such that the line through $P$ and $M$ is parallel to the $x$-axis\\
(b) determine an expression for\\
(i) $P S ^ { 2 }$ in terms of $e , x$ and $y$\\
(ii) $P M ^ { 2 }$ in terms of $e$ and $x$\\
(c) Hence show that
\end{itemize}
$$b ^ { 2 } = 49 \left( 1 - e ^ { 2 } \right)$$
Given that $E$ crosses the $y$-axis at the points with coordinates $( 0 , \pm 4 \sqrt { 3 } )$\\
(d) determine the value of $e$
Given that the $x$ coordinate of $P$ is $\frac { 7 } { 2 }$\\
(e) determine the area of triangle $O P M$, where $O$ is the origin.
\hfill \mbox{\textit{Edexcel F3 2024 Q3 [11]}}