| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola focus-directrix properties |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring deep understanding of hyperbola properties (eccentricity, foci, directrices, asymptotes), coordinate geometry to find intersection points, area calculations, and algebraic manipulation to derive and solve a cubic equation. While systematic, it demands multiple sophisticated techniques and extended reasoning across four connected parts, placing it well above average difficulty but not at the extreme end since the steps are guided. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^24.02c Complex notation: z, z*, Re(z), Im(z), |z|, arg(z) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2 = a^2(e^2-1) \Rightarrow e^2 = \frac{25}{a^2}+1 = \frac{25+a^2}{a^2}\) oe | B1 | Correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = (\pm)\frac{a}{e}\) or \(\frac{x}{a} = (\pm)\frac{y}{5}\) | B1 | |
| \(\frac{a}{e}\times\frac{1}{a} = \pm\frac{y}{5} \Rightarrow y = \pm\frac{5}{e} \Rightarrow AA' = 2\times\frac{5}{e}\) or \(\frac{5}{e}-\left(-\frac{5}{e}\right)\) | M1 | |
| \(= \frac{10}{e}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}\times\text{"}\frac{10}{e}\text{"}\times\left(ae+\frac{a}{e}\right)\) or e.g. \(\frac{1}{2}\times\frac{10a}{\sqrt{25+a^2}}\times\left(\sqrt{25+a^2}+\frac{a^2}{\sqrt{25+a^2}}\right)\) | M1 | |
| \(\frac{1}{2}\cdot\frac{10}{e}\left(ae+\frac{a}{e}\right) = \frac{164}{3} \Rightarrow 15\left(a+\frac{a}{e^2}\right) = 164\) | M1 | |
| \(\Rightarrow 15a\left(1+\frac{a^2}{25+a^2}\right) = 164\) | A1 (M1 on EPEN) | |
| \(\Rightarrow 15a\left(\frac{25+2a^2}{25+a^2}\right) = 164 \Rightarrow 375a+30a^3 = 164(25+a^2)\) \(\Rightarrow 30a^3 - 164a^2 + 375a - 4100 = 0\) * | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(30a^3 - 164a^2 + 375a - 4100 = (3a-20)(10a^2+12a+205)\) | B1 (M1 on EPEN) | |
| \(12^2 - 4(10)(205) = \ldots\) or \(10a^2+12a+205 = 10\left(\left(a+\frac{12}{20}\right)^2 - \frac{144}{400}\right)+205\) | M1 | |
| \(12^2 - 4(10)(205) < 0\) so no other real roots; hence \(a = \frac{20}{3}\) is only possible value | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies at least one correct directrix equation and at least one asymptote, including \(b=5\) | B1 | Stated or used |
| Solves to find \(y\)-coordinates of \(A\) and \(A'\), or just one and doubles to get length | M1 | Allow if \(b\) used rather than 5 |
| Correct length (from subtracting or doubling), must be positive | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses focus \((-ae, 0)\) and directrix \(x = \frac{a}{e}\) with length from (b) to form correct expression for area of triangle \(AFA'\) | M1 | Allow alternative pair |
| Sets area equation equal to \(\frac{164}{3}\) to obtain equation in \(e^2\) and \(a\) | M1 | Area must be of form \(\frac{1}{2} \times \frac{"10"}{e} \times \pm\left(ae \pm \frac{a}{e}\right)\) |
| Correct equation in terms of \(a\) only | A1(M1 on EPEN) | Allow any correct form |
| Correct result with no errors and sufficient working | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Shows \(a = \frac{20}{3}\) is a solution, e.g. \(30a^3 - 164a^2 + 375a - 4100 = (3a-20)(10a^2+12a+205)\) or \(f\!\left(\frac{20}{3}\right) = \frac{80000}{9} - \frac{65600}{9} + 2500 - 4100 = 0\) | B1(M1 on EPEN) | Or long division with correct quotient and no remainder |
| Correct method showing no other roots, e.g. \(\frac{d}{da}(eqn) = 90a^2 - 328a + 375 = 90\!\left(a - \frac{82}{45}\right)^2 + \frac{3427}{45} > 0\) so strictly increasing, hence only one solution | M1 | May use completing the square, discriminant, or differentiation. If discriminant used, values must be evaluated |
| \(a = \frac{20}{3}\) is the only possible value, all work correct with reason and conclusion | A1 | Discriminant must be correct, e.g. \(12^2 - 4(10)(205) = -8056\). Note: just using a calculator to solve the cubic scores no marks |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2 = a^2(e^2-1) \Rightarrow e^2 = \frac{25}{a^2}+1 = \frac{25+a^2}{a^2}$ oe | B1 | Correct expression |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = (\pm)\frac{a}{e}$ or $\frac{x}{a} = (\pm)\frac{y}{5}$ | B1 | |
| $\frac{a}{e}\times\frac{1}{a} = \pm\frac{y}{5} \Rightarrow y = \pm\frac{5}{e} \Rightarrow AA' = 2\times\frac{5}{e}$ or $\frac{5}{e}-\left(-\frac{5}{e}\right)$ | M1 | |
| $= \frac{10}{e}$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\times\text{"}\frac{10}{e}\text{"}\times\left(ae+\frac{a}{e}\right)$ or e.g. $\frac{1}{2}\times\frac{10a}{\sqrt{25+a^2}}\times\left(\sqrt{25+a^2}+\frac{a^2}{\sqrt{25+a^2}}\right)$ | M1 | |
| $\frac{1}{2}\cdot\frac{10}{e}\left(ae+\frac{a}{e}\right) = \frac{164}{3} \Rightarrow 15\left(a+\frac{a}{e^2}\right) = 164$ | M1 | |
| $\Rightarrow 15a\left(1+\frac{a^2}{25+a^2}\right) = 164$ | A1 (M1 on EPEN) | |
| $\Rightarrow 15a\left(\frac{25+2a^2}{25+a^2}\right) = 164 \Rightarrow 375a+30a^3 = 164(25+a^2)$ $\Rightarrow 30a^3 - 164a^2 + 375a - 4100 = 0$ * | A1* | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $30a^3 - 164a^2 + 375a - 4100 = (3a-20)(10a^2+12a+205)$ | B1 (M1 on EPEN) | |
| $12^2 - 4(10)(205) = \ldots$ or $10a^2+12a+205 = 10\left(\left(a+\frac{12}{20}\right)^2 - \frac{144}{400}\right)+205$ | M1 | |
| $12^2 - 4(10)(205) < 0$ so no other real roots; hence $a = \frac{20}{3}$ is only possible value | A1 | |
# Question (b) — Hyperbola/Conic Section:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies at least one correct directrix equation and at least one asymptote, including $b=5$ | B1 | Stated or used |
| Solves to find $y$-coordinates of $A$ and $A'$, or just one and doubles to get length | M1 | Allow if $b$ used rather than 5 |
| Correct length (from subtracting or doubling), must be positive | A1 | |
# Question (c) — Area of Triangle:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses focus $(-ae, 0)$ and directrix $x = \frac{a}{e}$ with length from (b) to form correct expression for area of triangle $AFA'$ | M1 | Allow alternative pair |
| Sets area equation equal to $\frac{164}{3}$ to obtain equation in $e^2$ and $a$ | M1 | Area must be of form $\frac{1}{2} \times \frac{"10"}{e} \times \pm\left(ae \pm \frac{a}{e}\right)$ |
| Correct equation in terms of $a$ only | A1(M1 on EPEN) | Allow any correct form |
| Correct result with no errors and sufficient working | A1* | |
# Question (d) — Cubic Root:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Shows $a = \frac{20}{3}$ is a solution, e.g. $30a^3 - 164a^2 + 375a - 4100 = (3a-20)(10a^2+12a+205)$ or $f\!\left(\frac{20}{3}\right) = \frac{80000}{9} - \frac{65600}{9} + 2500 - 4100 = 0$ | B1(M1 on EPEN) | Or long division with correct quotient and no remainder |
| Correct method showing no other roots, e.g. $\frac{d}{da}(eqn) = 90a^2 - 328a + 375 = 90\!\left(a - \frac{82}{45}\right)^2 + \frac{3427}{45} > 0$ so strictly increasing, hence only one solution | M1 | May use completing the square, discriminant, or differentiation. If discriminant used, values must be evaluated |
| $a = \frac{20}{3}$ is the only possible value, all work correct with reason and conclusion | A1 | Discriminant must be correct, e.g. $12^2 - 4(10)(205) = -8056$. Note: just using a calculator to solve the cubic scores no marks |
---7. A hyperbola $H$ has equation
$$\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { 25 } = 1$$
where $a$ is a positive constant.\\
The eccentricity of $H$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item Determine an expression for $e ^ { 2 }$ in terms of $a$.
The line $l$ is the directrix of $H$ for which $x > 0$\\
The points $A$ and $A ^ { \prime }$ are the points of intersection of $l$ with the asymptotes of $H$.
\item Determine, in terms of $e$, the length of the line segment $A A ^ { \prime }$.
The point $F$ is the focus of $H$ for which $x < 0$\\
Given that the area of triangle $A F A ^ { \prime }$ is $\frac { 164 } { 3 }$
\item show that $a$ is a solution of the equation
$$30 a ^ { 3 } - 164 a ^ { 2 } + 375 a - 4100 = 0$$
\item Hence, using algebra and making your reasoning clear, show that the only possible value of $a$ is $\frac { 20 } { 3 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2021 Q7 [11]}}