| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| 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 FP1 question requiring knowledge of ellipse and hyperbola properties (eccentricity, foci relationships). While it involves multiple steps and coordinate geometry of conics, the solution follows a systematic approach using standard formulas: find ellipse eccentricity and foci, apply the reciprocal relationship, then use the focus formula for hyperbolas to solve for a and b. The conceptual demand is moderate for FM students who know the formulas, but higher than typical A-level due to the topic itself. |
| Spec | 4.05c Partial fractions: extended to quadratic denominators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(9 = 25(1-e^2) \Rightarrow e = \ldots\left\{\frac{4}{5}\right\}\) | M1 | Complete method to find eccentricity of \(E\). Uses \(b^2 = a^2(1-e^2)\) to find \(e\) or \(e^2\). Ignore \(\pm\) references. |
| \(ae = 5 \times \frac{4}{5} = \ldots\{4\}\) | dM1 | Dependent on previous M. Finds \(x\)-coordinate of focus of \(E\), 5 multiplied by their \(e\). |
| \(e = \frac{1}{\text{their } \frac{4}{5}} = \ldots\left\{\frac{5}{4}\right\}\) and uses \(a = \frac{\text{their '4'}}{\text{their } \frac{5}{4}} = \ldots\left\{\frac{16}{5}\right\}\) | M1 | Complete method to find \(a\). Finds reciprocal of eccentricity and uses \(x\)-coordinate of focus divided by reciprocal of \(e\). |
| \(a = \frac{16}{5}\) or \(b = \frac{12}{5}\) | A1 | Correct value of \(a\) or \(b\). Accept as decimal. |
| Uses \(b^2 = \left(\text{their } \frac{16}{5}\right)^2\left(\left(\text{their } \frac{5}{4}\right)^2 - 1\right)\) leading to \(b = \ldots\) | M1 | Complete method to find \(b\). Uses \(b^2 = a^2(e^2-1)\) with their values. |
| \(a = \frac{16}{5}\) and \(b = \frac{12}{5}\) | A1 | Both correct. Accept as decimals. |
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $9 = 25(1-e^2) \Rightarrow e = \ldots\left\{\frac{4}{5}\right\}$ | **M1** | Complete method to find eccentricity of $E$. Uses $b^2 = a^2(1-e^2)$ to find $e$ or $e^2$. Ignore $\pm$ references. |
| $ae = 5 \times \frac{4}{5} = \ldots\{4\}$ | **dM1** | Dependent on previous M. Finds $x$-coordinate of focus of $E$, 5 multiplied by their $e$. |
| $e = \frac{1}{\text{their } \frac{4}{5}} = \ldots\left\{\frac{5}{4}\right\}$ and uses $a = \frac{\text{their '4'}}{\text{their } \frac{5}{4}} = \ldots\left\{\frac{16}{5}\right\}$ | **M1** | Complete method to find $a$. Finds reciprocal of eccentricity and uses $x$-coordinate of focus divided by reciprocal of $e$. |
| $a = \frac{16}{5}$ or $b = \frac{12}{5}$ | **A1** | Correct value of $a$ or $b$. Accept as decimal. |
| Uses $b^2 = \left(\text{their } \frac{16}{5}\right)^2\left(\left(\text{their } \frac{5}{4}\right)^2 - 1\right)$ leading to $b = \ldots$ | **M1** | Complete method to find $b$. Uses $b^2 = a^2(e^2-1)$ with their values. |
| $a = \frac{16}{5}$ and $b = \frac{12}{5}$ | **A1** | Both correct. Accept as decimals. |
---\begin{enumerate}
\item The ellipse $E$ has equation
\end{enumerate}
$$\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1$$
The hyperbola $H$ has equation
$$\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$$
where $a$ and $b$ are positive constants.\\
Given that
\begin{itemize}
\item the eccentricity of $H$ is the reciprocal of the eccentricity of $E$
\item the coordinates of the foci of $H$ are the same as the coordinates of the foci of $E$ determine\\
(i) the value of $a$\\
(ii) the value of $b$
\end{itemize}
\hfill \mbox{\textit{Edexcel FP1 2024 Q6 [6]}}