| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola focus-directrix properties |
| Difficulty | Standard +0.8 This is a Further Maths question on hyperbola focus-directrix properties requiring knowledge of the relationship e = c/a and directrix x = a/e. While the calculations are straightforward once the formulas are known, this topic is specialized (FM Pure 3) and requires understanding of eccentricity relationships not covered in standard A-level. The multi-step nature and FM content places it moderately above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = -\frac{4}{3}\) | B1 | \(x = -\frac{4}{3}\) or any equivalent equation. Allow \(x = \pm\frac{4}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{a}{e} = \frac{4}{3}\); \(b^2 = a^2(e^2-1) \Rightarrow 5 = a^2\left(\frac{9a^2}{16}-1\right)\) | M1 | Uses \(\frac{a}{e} = \pm\frac{4}{3}\) oe and a correct eccentricity formula, obtains equation in \(a\). Condone replacing \(b^2\) with 25 if equation is otherwise correct |
| \(9a^4 - 16a^2 - 80 = 0 \Rightarrow (9a^2+20)(a^2-4) = 0 \Rightarrow a^2 = \ldots\) | dM1 | Solves 3TQ in \(a^2\) to find positive real root. Must obtain one positive real value of \(a^2\) or \(a\) correct to 3sf consistent with their equation. Requires previous M mark |
| \(a = 2\) | A1 | Not \(a = \pm 2\) unless negative rejected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{a}{e} = \frac{4}{3}\); \(b^2 = a^2(e^2-1) \Rightarrow 5 = \left(\frac{4e}{3}\right)^2(e^2-1)\) | M1 | Uses \(\frac{a}{e} = \pm\frac{4}{3}\) oe and correct eccentricity formula, obtains equation in \(e\). Condone replacing \(b^2\) with 25 |
| \(16e^4 - 16e^2 - 45 = 0 \Rightarrow (4e^2-9)(4e^2+5) = 0 \Rightarrow e^2 = \ldots\) | dM1 | Solves 3TQ in \(e^2\) to find positive real root. Requires previous M mark |
| \(\left(e = \frac{3}{2} \Rightarrow\right)\quad a = 2\) | A1 | Not \(a = \pm2\) unless negative rejected; condone sight of \(e = \pm\frac{3}{2}\) or \(e = -\frac{3}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e = \frac{3}{2} \Rightarrow ae = \frac{3}{2}\times 2\) or \(ae = \frac{3a^2}{4} = \frac{3}{4}\times 4\) or \(ae = c = \sqrt{a^2+b^2} = \sqrt{2^2+5}\) | M1 | Uses correct method to obtain numerical expression for \(ae\) with their values of \(a\), \(e\), \(a^2\), \(b^2\). Condone use of negative \(e\) or \(a\) |
| Foci are \((\pm 3, 0)\) | A1 | Both correct foci as coordinates |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = -\frac{4}{3}$ | B1 | $x = -\frac{4}{3}$ or any equivalent equation. Allow $x = \pm\frac{4}{3}$ |
## Part (b)(i) — Way 1
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{a}{e} = \frac{4}{3}$; $b^2 = a^2(e^2-1) \Rightarrow 5 = a^2\left(\frac{9a^2}{16}-1\right)$ | M1 | Uses $\frac{a}{e} = \pm\frac{4}{3}$ oe and a correct eccentricity formula, obtains equation in $a$. Condone replacing $b^2$ with 25 if equation is otherwise correct |
| $9a^4 - 16a^2 - 80 = 0 \Rightarrow (9a^2+20)(a^2-4) = 0 \Rightarrow a^2 = \ldots$ | dM1 | Solves 3TQ in $a^2$ to find positive real root. Must obtain one positive real value of $a^2$ or $a$ correct to 3sf consistent with their equation. Requires previous M mark |
| $a = 2$ | A1 | Not $a = \pm 2$ unless negative rejected |
## Part (b)(i) — Way 2
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{a}{e} = \frac{4}{3}$; $b^2 = a^2(e^2-1) \Rightarrow 5 = \left(\frac{4e}{3}\right)^2(e^2-1)$ | M1 | Uses $\frac{a}{e} = \pm\frac{4}{3}$ oe and correct eccentricity formula, obtains equation in $e$. Condone replacing $b^2$ with 25 |
| $16e^4 - 16e^2 - 45 = 0 \Rightarrow (4e^2-9)(4e^2+5) = 0 \Rightarrow e^2 = \ldots$ | dM1 | Solves 3TQ in $e^2$ to find positive real root. Requires previous M mark |
| $\left(e = \frac{3}{2} \Rightarrow\right)\quad a = 2$ | A1 | Not $a = \pm2$ unless negative rejected; condone sight of $e = \pm\frac{3}{2}$ or $e = -\frac{3}{2}$ |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e = \frac{3}{2} \Rightarrow ae = \frac{3}{2}\times 2$ or $ae = \frac{3a^2}{4} = \frac{3}{4}\times 4$ or $ae = c = \sqrt{a^2+b^2} = \sqrt{2^2+5}$ | M1 | Uses correct method to obtain numerical expression for $ae$ with their values of $a$, $e$, $a^2$, $b^2$. Condone use of negative $e$ or $a$ |
| Foci are $(\pm 3, 0)$ | A1 | Both correct foci as coordinates |
---\begin{enumerate}
\item A hyperbola $H$ has equation
\end{enumerate}
$$\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { 5 } = 1 \quad \text { where } a \text { is a positive constant }$$
The line with equation $x = \frac { 4 } { 3 }$ is a directrix of $H$\\
(a) Write down an equation of the other directrix.\\
(b) Determine\\
(i) the value of $a$\\
(ii) the coordinates of each of the foci of $H$
\hfill \mbox{\textit{Edexcel F3 2023 Q2 [6]}}