| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic tangent through external point |
| Difficulty | Challenging +1.2 This is a structured multi-part question on ellipse properties and tangency conditions. Part (a) requires standard focus-directrix relationships, (b) is algebraic substitution, (c) uses the discriminant condition for tangency (a standard technique), (d) involves basic coordinate geometry, and (e) requires calculus optimization. While it spans multiple techniques and requires careful algebra, each step follows predictable Further Maths methods without requiring novel insight. The scaffolding guides students through the solution, making it moderately above average difficulty but well within reach for prepared FP3 students. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.07n Stationary points: find maxima, minima using derivatives |
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| Q7 | |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(ae = 3\), \(\frac{a}{e} = \frac{25}{3}\) or \(ae = \pm3\), \(\frac{a}{e} = \pm\frac{25}{3}\) | B1 | Correct equations. Ignore use of \(+\) or \(-\) throughout |
| \(e^2 = \frac{9}{25}\) and \(a^2 = 25\) | M1 | Solves to find \(a\) or \(a^2\) and \(e\) or \(e^2\) |
| \(b^2 = a^2(1-e^2) \Rightarrow b^2 = 25\left(1 - \frac{9}{25}\right) = 16\) | M1 | Uses correct eccentricity formula to find \(b\) or \(b^2\) |
| \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \Rightarrow \frac{x^2}{25} + \frac{y^2}{16} = 1\) \(\left(\text{or } \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1\right)\) | M1A1 | M1: Uses correct ellipse formula with their \(a\) and \(b\). A1: Correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\frac{x^2}{25} + \frac{(mx+c)^2}{16} = 1\) | M1 | Substitutes for \(y\). Allow in terms of \(a\) and \(b\) |
| \(16x^2 + 25(m^2x^2 + 2mcx + c^2) = 400\) \(\therefore (16 + 25m^2)x^2 + 50mcx + 25(c^2 - 16) = 0\) | A1* | Correct proof including sufficient intermediate working (at least one step) with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(b^2 - 4ac = 0 \Rightarrow (50mc)^2 - 4(16+25m^2)(25(c^2-16)) = 0\) | M1A1 | M1: Uses \(b^2 - 4ac = 0\) with the given quadratic (may be implied). Do not allow as part of attempt to use quadratic formula unless discriminant is "extracted" and used \(= 0\). A1: Correct equation (the "\(= 0\)" may be implied/appear later) |
| \(c^2 = 25m^2 + 16\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(x = \pm\frac{\sqrt{25m^2+16}}{m}\), \(y = \sqrt{25m^2+16}\) | B1ft | Follow through their \(p\) and \(q\). May be implied by their attempt at the triangle area |
| Area \(OAB\ (=T) = \frac{1}{2} \cdot \frac{\sqrt{25m^2+16}}{m} \cdot \sqrt{25m^2+16}\) | M1 | Correct triangle area method (Allow \(\pm\) area here) |
| \(T = \frac{25m^2+16}{2m}\) * | A1* | Correct area. (Must be positive) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(y = mx + c \Rightarrow y = c\), \(x = \pm\frac{c}{m}\) | B1 | Correct intercepts |
| Area \(OAB\ (=T) = \frac{1}{2} \times c \times \frac{c}{m} = \frac{c^2}{2m}\) | M1 | Correct triangle area method (Allow \(\pm\) area here) |
| \(T = \frac{25m^2+16}{2m}\) * | A1* | Correct positive area. Must follow the final A1 in part (c) unless the work for part (c) is done in part (d) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(\frac{dT}{dm} = \frac{25}{2} - \frac{8}{m^2} = 0 \Rightarrow m = \frac{4}{5}\) or \(\frac{dT}{dm} = \frac{2m(50m) - 2(25m^2+16)}{4m^2} = 0 \Rightarrow m = \frac{4}{5}\) | M1 | Solves \(\frac{dT}{dm} = 0\) to obtain a value for \(m\) |
| \(m = \frac{4}{5} \Rightarrow T = 20\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(T = \frac{25m^2+16}{2m} = \frac{(5m-4)^2 + 40m}{2m}\), \((5m-4)^2 = 0 \Rightarrow T = \frac{40m}{2m}\). Writes \(T\) as \(\frac{(5m-4)^2 + \ldots}{2m}\) and realises minimum when \((5m-4)=0\) | M1 | |
| \(T = 20\) | A1 | cao |
# Question 7:
## Part (a):
| Working | Mark | Notes |
|---------|------|-------|
| $ae = 3$, $\frac{a}{e} = \frac{25}{3}$ or $ae = \pm3$, $\frac{a}{e} = \pm\frac{25}{3}$ | B1 | Correct equations. Ignore use of $+$ or $-$ throughout |
| $e^2 = \frac{9}{25}$ and $a^2 = 25$ | M1 | Solves to find $a$ or $a^2$ and $e$ or $e^2$ |
| $b^2 = a^2(1-e^2) \Rightarrow b^2 = 25\left(1 - \frac{9}{25}\right) = 16$ | M1 | Uses correct eccentricity formula to find $b$ or $b^2$ |
| $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \Rightarrow \frac{x^2}{25} + \frac{y^2}{16} = 1$ $\left(\text{or } \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1\right)$ | M1A1 | M1: Uses correct ellipse formula with their $a$ and $b$. A1: Correct equation |
**Total: (5)**
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## Part (b):
| Working | Mark | Notes |
|---------|------|-------|
| $\frac{x^2}{25} + \frac{(mx+c)^2}{16} = 1$ | M1 | Substitutes for $y$. Allow in terms of $a$ and $b$ |
| $16x^2 + 25(m^2x^2 + 2mcx + c^2) = 400$ $\therefore (16 + 25m^2)x^2 + 50mcx + 25(c^2 - 16) = 0$ | A1* | Correct proof including sufficient intermediate working (at least one step) with no errors |
**Total: (2)**
---
## Part (c):
| Working | Mark | Notes |
|---------|------|-------|
| $b^2 - 4ac = 0 \Rightarrow (50mc)^2 - 4(16+25m^2)(25(c^2-16)) = 0$ | M1A1 | M1: Uses $b^2 - 4ac = 0$ with the given quadratic (may be implied). Do not allow as part of attempt to use quadratic formula unless discriminant is "extracted" and used $= 0$. A1: Correct equation (the "$= 0$" may be implied/appear later) |
| $c^2 = 25m^2 + 16$ | A1 | cao |
**Total: (3)**
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## Part (d):
| Working | Mark | Notes |
|---------|------|-------|
| $x = \pm\frac{\sqrt{25m^2+16}}{m}$, $y = \sqrt{25m^2+16}$ | B1ft | Follow through their $p$ and $q$. May be implied by their attempt at the triangle area |
| Area $OAB\ (=T) = \frac{1}{2} \cdot \frac{\sqrt{25m^2+16}}{m} \cdot \sqrt{25m^2+16}$ | M1 | Correct triangle area method (Allow $\pm$ area here) |
| $T = \frac{25m^2+16}{2m}$ * | A1* | Correct area. (Must be positive) |
**Alternative (d):**
| Working | Mark | Notes |
|---------|------|-------|
| $y = mx + c \Rightarrow y = c$, $x = \pm\frac{c}{m}$ | B1 | Correct intercepts |
| Area $OAB\ (=T) = \frac{1}{2} \times c \times \frac{c}{m} = \frac{c^2}{2m}$ | M1 | Correct triangle area method (Allow $\pm$ area here) |
| $T = \frac{25m^2+16}{2m}$ * | A1* | Correct positive area. Must follow the final A1 in part (c) unless the work for part (c) is done in part (d) |
**Total: (3)**
---
## Part (e):
| Working | Mark | Notes |
|---------|------|-------|
| $\frac{dT}{dm} = \frac{25}{2} - \frac{8}{m^2} = 0 \Rightarrow m = \frac{4}{5}$ **or** $\frac{dT}{dm} = \frac{2m(50m) - 2(25m^2+16)}{4m^2} = 0 \Rightarrow m = \frac{4}{5}$ | M1 | Solves $\frac{dT}{dm} = 0$ to obtain a value for $m$ |
| $m = \frac{4}{5} \Rightarrow T = 20$ | A1 | cao |
**Alternative (e):**
| Working | Mark | Notes |
|---------|------|-------|
| $T = \frac{25m^2+16}{2m} = \frac{(5m-4)^2 + 40m}{2m}$, $(5m-4)^2 = 0 \Rightarrow T = \frac{40m}{2m}$. Writes $T$ as $\frac{(5m-4)^2 + \ldots}{2m}$ and realises minimum when $(5m-4)=0$ | M1 | |
| $T = 20$ | A1 | cao |
**Total: (2)**
**Question Total: 15**7. The ellipse $E$ has foci at the points $( \pm 3,0 )$ and has directrices with equations $x = \pm \frac { 25 } { 3 }$
\begin{enumerate}[label=(\alph*)]
\item Find a cartesian equation for the ellipse $E$.
The straight line $l$ has equation $y = m x + c$, where $m$ and $c$ are positive constants.
\item Show that the $x$ coordinates of any points of intersection of $l$ and $E$ satisfy the equation
$$\left( 16 + 25 m ^ { 2 } \right) x ^ { 2 } + 50 m c x + 25 \left( c ^ { 2 } - 16 \right) = 0$$
Given that the line $l$ is a tangent to $E$,
\item show that $c ^ { 2 } = p m ^ { 2 } + q$, where $p$ and $q$ are constants to be found.
The line $l$ intersects the $x$-axis at the point $A$ and intersects the $y$-axis at the point $B$.
\item Show that the area of triangle $O A B$, where $O$ is the origin, is
$$\frac { 25 m ^ { 2 } + 16 } { 2 m }$$
\item Find the minimum area of triangle $O A B$.
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\hfill \mbox{\textit{Edexcel FP3 2018 Q7 [15]}}