Edexcel F3 2014 June — Question 2 6 marks

Exam BoardEdexcel
ModuleF3 (Further Pure Mathematics 3)
Year2014
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConic sections
TypeEllipse focus-directrix properties
DifficultyChallenging +1.2 This is a Further Maths question requiring knowledge of the focus-directrix property of ellipses (directrix at x = a²/ae where e is eccentricity) and the relationship b² = a²(1-e²). Students must set up and solve a quadratic equation, but the conceptual framework is standard for F3. More challenging than typical A-level due to the Further Maths content, but straightforward application of known formulas.
Spec1.03d Circles: equation (x-a)^2+(y-b)^2=r^2

2. The line with equation \(x = 9\) is a directrix of an ellipse with equation $$\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { 8 } = 1$$ where \(a\) is a positive constant. Find the two possible exact values of the constant \(a\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\pm\frac{a}{e} = \pm 9\) and \(a^2(1-e^2)=8\)B1 Both equations correct
\(a^4 - 81a^2 + 648 = 0\) or \(81e^4 - 81e^2 + 8 = 0\)M1A1 M1: Eliminates an unknown to produce a quadratic in \(a^2\) or \(e^2\). A1: Correct three term quadratic in any form with terms collected
\((a^2-72)(a^2-9)=0 \Rightarrow a^2 = \ldots\) or \((9e^2-8)(9e^2-1)=0 \Rightarrow e^2 = \ldots\)M1 Uses standard method to solve quadratic as far as \(a^2=\ldots\) or \(e^2=\ldots\) (Must be \(a^2=\ldots\) or \(e^2=\ldots\) at this stage). May be implied by correct answers only
\(a=3\) and \(a=6\sqrt{2}\)M1A1 M1: Complete method to find \(a\); either square roots from \(a^2=\ldots\) or square roots from \(e^2=\ldots\) and uses \(a=9e\) at least once. A1: cao (both answers correct). Do not accept \(\pm\) for either answer unless negative is rejected later
# Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pm\frac{a}{e} = \pm 9$ **and** $a^2(1-e^2)=8$ | B1 | Both equations correct |
| $a^4 - 81a^2 + 648 = 0$ or $81e^4 - 81e^2 + 8 = 0$ | M1A1 | M1: Eliminates an unknown to produce a quadratic in $a^2$ or $e^2$. A1: Correct three term quadratic in any form with terms collected |
| $(a^2-72)(a^2-9)=0 \Rightarrow a^2 = \ldots$ or $(9e^2-8)(9e^2-1)=0 \Rightarrow e^2 = \ldots$ | M1 | Uses standard method to solve quadratic as far as $a^2=\ldots$ or $e^2=\ldots$ (Must be $a^2=\ldots$ or $e^2=\ldots$ at this stage). May be implied by correct answers only |
| $a=3$ and $a=6\sqrt{2}$ | M1A1 | M1: Complete method to find $a$; either square roots from $a^2=\ldots$ or square roots from $e^2=\ldots$ and uses $a=9e$ at least once. A1: cao (both answers correct). Do not accept $\pm$ for either answer unless negative is rejected later |

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2. The line with equation $x = 9$ is a directrix of an ellipse with equation

$$\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { 8 } = 1$$

where $a$ is a positive constant.

Find the two possible exact values of the constant $a$.\\

\hfill \mbox{\textit{Edexcel F3 2014 Q2 [6]}}