| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse locus problems |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths ellipse question requiring standard techniques: finding foci/directrices from standard form, verifying the focal distance property (routine application of ellipse definition), and proving a locus result using coordinate geometry. Part (c) requires algebraic manipulation but follows a well-established method (chord midpoint locus). While more demanding than typical A-level questions due to the Further Maths content, it's largely procedural without requiring novel geometric insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2 = a^2(1-e^2) \Rightarrow 1 = 9(1-e^2)\) | M1 | Uses correct eccentricity formula with correct values for \(a\) and \(b\), obtains a value for \(e^2\) or \(e\) |
| \(e^2 = \ldots\left(\frac{8}{9}\right),\ e = \frac{2\sqrt{2}}{3}\) or \(\frac{\sqrt{8}}{3}\) | A1 | Correct value for \(e\) (not \(\pm\)). Could be implied |
| Foci are \(\left(\pm 2\sqrt{2}, 0\right)\) or \(\left(\pm\sqrt{8}, 0\right)\) | B1 | Both correct foci as coordinates. Condone any use of negative \(e\). Note: not an ft mark |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \pm\frac{9\sqrt{2}}{4}\) or \(\pm\frac{9\sqrt{8}}{8}\) or \(\pm\frac{9}{\sqrt{8}}\) | B1ft | Both correct equations. Requires single fraction. Allow ft: \(x = \pm\frac{3}{\text{their }e}\) computed into single fraction, condoning \(e<0\). Allow "\(x_1 = \ldots, x_2 = \ldots\)". Condone \("x = \pm\frac{a}{e} = \frac{9\sqrt{2}}{4}\) or \(-\frac{9\sqrt{2}}{4}"\) but just \("\frac{a}{e} = \pm\frac{9\sqrt{2}}{4}"\) is B0 |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | PF_1 | = e |
| \( | PF_1 | + |
| \(= 6\) | A1* | Fully correct proof. Modulus signs not required |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | PF_1 | + |
| e.g. \(Q\) where \(E\) crosses positive \(x\)-axis \(\Rightarrow | PF_1 | + |
| \(= 6\) | A1* | Fully correct proof |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(3\cos\theta, \sin\theta)\); \( | PF_1 | ^2 = (3\cos\theta - "2\sqrt{2}")^2 + \sin^2\theta\) or \( |
| \( | PF_1 | + |
| \(= 3-2\sqrt{2}\cos\theta + 3 + 2\sqrt{2}\cos\theta = 6\) | A1* | Fully correct proof. Intermediate step shown is required for this Way |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P\!\left(x,\sqrt{1-\frac{x^2}{9}}\right)\) or \(P\!\left(x,\sqrt{\frac{9-x^2}{9}}\right)\); \( | PF_1 | ^2 = ("2\sqrt{2}"-x)^2+1-\frac{x^2}{9}\) |
| \( | PF_1 | + |
| \(= 3-\frac{2\sqrt{2}}{3}x+3+\frac{2\sqrt{2}}{3}x = 6\) | A1* | Fully correct proof. Intermediate step required |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 + 9(2x+c)^2 = 9\) or \(\frac{x^2}{9}+(2x+c)^2=1\) | M1 | Substitutes line into ellipse equation. Condone slips provided intention clear |
| \(37x^2+36cx+9c^2-9=0\) or e.g. \(\frac{37}{9}x^2+4cx+c^2-1=0\) | A1 | Correct quadratic with \(x^2\) terms collected (could be implied) |
| \(\frac{1}{2}(\text{sum of roots}) \Rightarrow x = \frac{-18c}{37}\) | dM1 A1 | M1: Correct attempt at \(\frac{1}{2}\)(sum of roots), i.e. \(-\frac{b}{2a}\) for their quadratic. Ignore labelling. Requires previous M mark. A1: Any correct equation in \(x\) and \(c\) |
| \(c = -"\frac{37}{18}"x \Rightarrow y = 2x+\left(-"\frac{37}{18}"\right)x\) or \(x = -"\frac{18}{37}"c \Rightarrow y = 2\times-"\frac{18}{37}"c+c\) | ddM1 | Substitutes \(c=px\) into line to get equation in \(x\) and \(y\) only. Must not use "\(M_x\)" or "\(M_y\)" implying locus equation. Requires both previous M marks |
| \(y_{\_} = -\frac{1}{18}x_{\_}\) oe \(\therefore l\) passes through origin \(O\) | A1* | Obtains correct locus equation and concludes e.g. passes through origin \(O(0,0)\). Allow "shown"/"as required"/"QED". Requires all previous marks |
| (6) | ||
| Total 13 | ||
| Paper Total: 75 |
# Question 9:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2 = a^2(1-e^2) \Rightarrow 1 = 9(1-e^2)$ | M1 | Uses correct eccentricity formula with correct values for $a$ and $b$, obtains a value for $e^2$ or $e$ |
| $e^2 = \ldots\left(\frac{8}{9}\right),\ e = \frac{2\sqrt{2}}{3}$ or $\frac{\sqrt{8}}{3}$ | A1 | Correct value for $e$ (not $\pm$). Could be implied |
| Foci are $\left(\pm 2\sqrt{2}, 0\right)$ or $\left(\pm\sqrt{8}, 0\right)$ | B1 | Both correct foci as coordinates. Condone any use of negative $e$. **Note: not an ft mark** |
| | **(3)** | |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \pm\frac{9\sqrt{2}}{4}$ or $\pm\frac{9\sqrt{8}}{8}$ or $\pm\frac{9}{\sqrt{8}}$ | B1ft | Both correct **equations**. Requires single fraction. Allow ft: $x = \pm\frac{3}{\text{their }e}$ computed into single fraction, condoning $e<0$. Allow "$x_1 = \ldots, x_2 = \ldots$". Condone $"x = \pm\frac{a}{e} = \frac{9\sqrt{2}}{4}$ or $-\frac{9\sqrt{2}}{4}"$ but just $"\frac{a}{e} = \pm\frac{9\sqrt{2}}{4}"$ is B0 |
| | **(2)** | |
## Part (b) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|PF_1| = e|PM_1|$ or $|PF_2| = e|PM_2|$ | M1 | States this definition of an ellipse |
| $|PF_1|+|PF_2| = e(|PM_1|+|PM_2|)$ or $e(|M_1M_2|)$; $"\frac{2\sqrt{2}}{3}" \times 2 \times "\frac{9\sqrt{2}}{4}"$ oe | dM1 | Correct method for a numerical expression (or with cancelling "$x$"s) for $|PF_1|+|PF_2|$ with their $e$ and directrix. **One of the underlined expressions must be seen for first approach. Requires previous M mark.** |
| $= 6$ | A1* | Fully correct proof. Modulus signs not required |
| | **(3)** | |
## Part (b) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|PF_1|+|PF_2| = |QF_1|+|QF_2|$ where $P$ and $Q$ any points on ellipse | M1 | States oe definition of ellipse, justified by explanation. Accept "$|PF_1|+|PF_2|$ is constant for any $P$" |
| e.g. $Q$ where $E$ crosses positive $x$-axis $\Rightarrow |PF_1|+|PF_2| = 3-"2\sqrt{2}"+3+"2\sqrt{2}"$ | dM1 | Correct method for numerical value using another point and foci. **Requires previous M mark.** |
| $= 6$ | A1* | Fully correct proof |
| | **(3)** | |
## Part (b) — Way 3 (Point in terms of $\theta$):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(3\cos\theta, \sin\theta)$; $|PF_1|^2 = (3\cos\theta - "2\sqrt{2}")^2 + \sin^2\theta$ or $|PF_2|^2 = (3\cos\theta + "2\sqrt{2}")^2 + \sin^2\theta$ | M1 | Correct general point in parametric form, applies Pythagoras for distance in terms of $a$, $b$ and $\theta$ |
| $|PF_1|+|PF_2| = \sqrt{8\cos^2\theta - 12\sqrt{2}\cos\theta+9} + \sqrt{8\cos^2\theta+12\sqrt{2}\cos\theta+9}$ | dM1 | Two three-term quadratic expressions required. **Requires previous M mark.** |
| $= 3-2\sqrt{2}\cos\theta + 3 + 2\sqrt{2}\cos\theta = 6$ | A1* | Fully correct proof. Intermediate step shown is required for this Way |
| | **(3)** | |
## Part (b) — Way 4 (Point in terms of $x$):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P\!\left(x,\sqrt{1-\frac{x^2}{9}}\right)$ or $P\!\left(x,\sqrt{\frac{9-x^2}{9}}\right)$; $|PF_1|^2 = ("2\sqrt{2}"-x)^2+1-\frac{x^2}{9}$ | M1 | Correct general point in terms of $x$, applies Pythagoras in terms of $a$, $b$ and $x$ |
| $|PF_1|+|PF_2| = \sqrt{\frac{8}{9}x^2-4\sqrt{2}x+9}+\sqrt{\frac{8}{9}x^2+4\sqrt{2}x+9}$ | dM1 | Two three-term quadratic expressions. **Requires previous M mark.** |
| $= 3-\frac{2\sqrt{2}}{3}x+3+\frac{2\sqrt{2}}{3}x = 6$ | A1* | Fully correct proof. Intermediate step required |
| | **(3)** | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 + 9(2x+c)^2 = 9$ or $\frac{x^2}{9}+(2x+c)^2=1$ | M1 | Substitutes line into ellipse equation. Condone slips provided intention clear |
| $37x^2+36cx+9c^2-9=0$ or e.g. $\frac{37}{9}x^2+4cx+c^2-1=0$ | A1 | Correct quadratic with $x^2$ terms collected (could be implied) |
| $\frac{1}{2}(\text{sum of roots}) \Rightarrow x = \frac{-18c}{37}$ | dM1 A1 | M1: Correct attempt at $\frac{1}{2}$(sum of roots), i.e. $-\frac{b}{2a}$ for their quadratic. Ignore labelling. **Requires previous M mark.** A1: Any correct **equation** in $x$ and $c$ |
| $c = -"\frac{37}{18}"x \Rightarrow y = 2x+\left(-"\frac{37}{18}"\right)x$ or $x = -"\frac{18}{37}"c \Rightarrow y = 2\times-"\frac{18}{37}"c+c$ | ddM1 | Substitutes $c=px$ into line to get equation in $x$ and $y$ only. Must not use "$M_x$" or "$M_y$" implying locus equation. **Requires both previous M marks** |
| $y_{\_} = -\frac{1}{18}x_{\_}$ oe $\therefore l$ passes through origin $O$ | A1* | Obtains correct locus equation **and** concludes e.g. passes through origin $O(0,0)$. Allow "shown"/"as required"/"QED". **Requires all previous marks** |
| | **(6)** | |
| | **Total 13** | |
| | **Paper Total: 75** | |\begin{enumerate}
\item The ellipse $E$ has equation
\end{enumerate}
$$x ^ { 2 } + 9 y ^ { 2 } = 9$$
The foci of $E$ are $F _ { 1 }$ and $F _ { 2 }$\\
(a) (i) Determine the coordinates of $F _ { 1 }$ and the coordinates of $F _ { 2 }$\\
(ii) Write down the equation of each of the directrices of $E$
The point $P$ lies on the ellipse.\\
(b) Show that $\left| P F _ { 1 } \right| + \left| P F _ { 2 } \right| = 6$
The straight line through $P$ with equation $y = 2 x + c$ meets $E$ again at the point $Q$ The point $M$ is the midpoint of $P Q$\\
(c) Show that as $P$ varies the locus of $M$ is a straight line passing through the origin.
\hfill \mbox{\textit{Edexcel F3 2023 Q9 [13]}}