| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola focus-directrix properties |
| Difficulty | Challenging +1.2 This is a multi-step Further Maths question requiring knowledge of eccentricity formulas for both ellipses and hyperbolas, plus algebraic manipulation. While it involves two conic sections and requires careful application of formulas (e² = 1 - b²/a² for ellipse, e² = 1 + b²/a² for hyperbola), the steps are fairly standard once the formulas are recalled. Part (a) is guided ('show that'), and part (b) follows directly from equating foci. Slightly above average difficulty due to being Further Maths content and requiring coordination of multiple formulas, but not requiring deep insight or novel problem-solving. |
| Spec | 4.03d Linear transformations 2D: reflection, rotation, enlargement, shear |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(b^2 = a^2(1-e_1^2) \Rightarrow 4 = 16(1-e_1^2) \Rightarrow e_1^2 = \ldots\) | M1 | Uses \(b^2 = a^2(1-e_1^2)\) with values for \(a\) and \(b\) to find a value for \(e_1\) or \(e_1^2\) |
| \(e_1^2 = \frac{3}{4}\) or \(e_1 = \frac{\sqrt{3}}{2}\) | A1 | Correct exact value for \(e_1\) or \(e_1^2\); allow M1A1 if relevant work seen in (b) |
| E.g. \(b^2 = a^2(e_2^2-1) = a^2\left(\frac{1}{e_1^2}-1\right) = a^2\left(\frac{4}{3}-1\right)\) | dM1 | Dependent on previous M1; uses \(e_1 \times e_2 = 1\) with \(e_1\) or \(e_1^2\) to find expression between \(a\) and \(b\) |
| \(\Rightarrow b^2 = \frac{1}{3}a^2 \Rightarrow a^2 = 3b^2\) * cso | A1* | Achieves \(a^2 = 3b^2\) with at least one intermediate unsimplified equation in \(a\) and \(b\); cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| For focus of ellipse: \((x=) 4 \times \frac{\sqrt{3}}{2}\) | M1 | Uses/implies \(x\) coordinate of focus for ellipse is \(4 \times e_1\) |
| For focus of hyperbola: \((x=) a \times \frac{2}{\sqrt{3}} \Rightarrow 2\sqrt{3} = \frac{2a}{\sqrt{3}} \Rightarrow a = \ldots (=3)\), then \(\Rightarrow b^2 = \frac{1}{3}a^2 = \ldots\) | M1 | Full process to find values for \(a\) and \(b\) or their squares; e.g. for focus of hyperbola \(x = a \times e_2 = \frac{a}{e_1}\), sets equal to \(4e_1\) and solves for \(a\) then uses \(a^2 = 3b^2\) to obtain \(b^2\) |
| \(\frac{x^2}{9} - \frac{y^2}{3} = 1\) cso | A1 | Deduces the correct equation for the hyperbola |
## Question 1:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $b^2 = a^2(1-e_1^2) \Rightarrow 4 = 16(1-e_1^2) \Rightarrow e_1^2 = \ldots$ | M1 | Uses $b^2 = a^2(1-e_1^2)$ with values for $a$ and $b$ to find a value for $e_1$ or $e_1^2$ |
| $e_1^2 = \frac{3}{4}$ or $e_1 = \frac{\sqrt{3}}{2}$ | A1 | Correct exact value for $e_1$ or $e_1^2$; allow M1A1 if relevant work seen in (b) |
| E.g. $b^2 = a^2(e_2^2-1) = a^2\left(\frac{1}{e_1^2}-1\right) = a^2\left(\frac{4}{3}-1\right)$ | dM1 | Dependent on previous M1; uses $e_1 \times e_2 = 1$ with $e_1$ or $e_1^2$ to find expression between $a$ and $b$ |
| $\Rightarrow b^2 = \frac{1}{3}a^2 \Rightarrow a^2 = 3b^2$ * cso | A1* | Achieves $a^2 = 3b^2$ with at least one intermediate unsimplified equation in $a$ and $b$; cso |
**SC:** Allow M0A0dM1A0 if $b^2 = a^2(1-e_1)$ and $b^2 = a^2(e_2-1)$ used in otherwise correct process.
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### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| For focus of ellipse: $(x=) 4 \times \frac{\sqrt{3}}{2}$ | M1 | Uses/implies $x$ coordinate of focus for ellipse is $4 \times e_1$ |
| For focus of hyperbola: $(x=) a \times \frac{2}{\sqrt{3}} \Rightarrow 2\sqrt{3} = \frac{2a}{\sqrt{3}} \Rightarrow a = \ldots (=3)$, then $\Rightarrow b^2 = \frac{1}{3}a^2 = \ldots$ | M1 | Full process to find values for $a$ and $b$ or their squares; e.g. for focus of hyperbola $x = a \times e_2 = \frac{a}{e_1}$, sets equal to $4e_1$ and solves for $a$ then uses $a^2 = 3b^2$ to obtain $b^2$ |
| $\frac{x^2}{9} - \frac{y^2}{3} = 1$ cso | A1 | Deduces the correct equation for the hyperbola |\begin{enumerate}
\item An ellipse has equation $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1$ and eccentricity $e _ { 1 }$ A hyperbola has equation $\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ and eccentricity $e _ { 2 }$
\end{enumerate}
Given that $e _ { 1 } \times e _ { 2 } = 1$\\
(a) show that $a ^ { 2 } = 3 b ^ { 2 }$
Given also that the coordinates of the foci of the ellipse are the same as the coordinates of the foci of the hyperbola,\\
(b) determine the equation of the hyperbola.
\hfill \mbox{\textit{Edexcel FP1 2022 Q1 [7]}}