| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola focus-directrix properties |
| Difficulty | Standard +0.8 This is a Further Maths FP3 question requiring knowledge of focus-directrix definition of conics and the relationship e = c/a = a/d for hyperbolas. Students must derive the eccentricity from given foci and directrices, then construct the standard form equation. While systematic, it requires understanding beyond standard A-level and involves algebraic manipulation across multiple steps, placing it moderately above average difficulty. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Foci \((\pm5, 0)\), Directrices \(x = \pm\frac{9}{5}\) | ||
| \((\pm)ae = (\pm)5\) and \((\pm)\frac{a}{e} = (\pm)\frac{9}{5}\) | B1 | Correct equations (ignore \(\pm\)'s) |
| \(e = \frac{5}{a} \Rightarrow \frac{a^2}{5} = \frac{9}{5} \Rightarrow a^2 = 9\) | M1 | Solves using an appropriate method to find \(a^2\) or \(a\) |
| or \(a = \frac{5}{e} \Rightarrow \frac{5}{e^2} = \frac{9}{5} \Rightarrow e = \frac{5}{3} \Rightarrow a = 3\) | A1 | \(a^2 = 9\) or \(a = (\pm)3\) |
| \(b^2 = a^2e^2 - a^2 \Rightarrow b^2 = 25 - 9\), so \(b^2 = 16\) \((\Rightarrow b=4)\) | M1 | Use of \(b^2 = a^2(e^2-1)\) to obtain a numerical value for \(b^2\) or \(b\) |
| or \(b^2 = a^2(e^2-1) \Rightarrow b^2 = 9\left(\frac{25}{9}-1\right)\), \(b^2 = 16\) \((\Rightarrow b=4)\) | A1 | \(b^2 = 16\) or \(b = (\pm)4\) |
| \(\frac{x^2}{9} - \frac{y^2}{16} = 1\) | M1 | Use of \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) with their \(a^2\) and \(b^2\) |
| A1 | Correct hyperbola in any form | |
| (7) |
# Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Foci $(\pm5, 0)$, Directrices $x = \pm\frac{9}{5}$ | | |
| $(\pm)ae = (\pm)5$ and $(\pm)\frac{a}{e} = (\pm)\frac{9}{5}$ | B1 | Correct equations (ignore $\pm$'s) |
| $e = \frac{5}{a} \Rightarrow \frac{a^2}{5} = \frac{9}{5} \Rightarrow a^2 = 9$ | M1 | Solves using an appropriate method to find $a^2$ or $a$ |
| or $a = \frac{5}{e} \Rightarrow \frac{5}{e^2} = \frac{9}{5} \Rightarrow e = \frac{5}{3} \Rightarrow a = 3$ | A1 | $a^2 = 9$ or $a = (\pm)3$ |
| $b^2 = a^2e^2 - a^2 \Rightarrow b^2 = 25 - 9$, so $b^2 = 16$ $(\Rightarrow b=4)$ | M1 | Use of $b^2 = a^2(e^2-1)$ to obtain a numerical value for $b^2$ or $b$ |
| or $b^2 = a^2(e^2-1) \Rightarrow b^2 = 9\left(\frac{25}{9}-1\right)$, $b^2 = 16$ $(\Rightarrow b=4)$ | A1 | $b^2 = 16$ or $b = (\pm)4$ |
| $\frac{x^2}{9} - \frac{y^2}{16} = 1$ | M1 | Use of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with their $a^2$ and $b^2$ |
| | A1 | Correct hyperbola in any form |
| | **(7)** | |
---\begin{enumerate}
\item The hyperbola $H$ has foci at $( 5,0 )$ and $( - 5,0 )$ and directrices with equations $x = \frac { 9 } { 5 }$ and $x = - \frac { 9 } { 5 }$.
\end{enumerate}
Find a cartesian equation for $H$.\\
\hfill \mbox{\textit{Edexcel FP3 2013 Q1 [7]}}