| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Hyperbola tangent and geometric properties |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on hyperbolas with parametric form using hyperbolic functions. Part (a) is routine implicit differentiation, part (b) is straightforward substitution, and part (c) requires showing perpendicularity using gradients and the focus location (c² = a² + b²). While it involves Further Maths content and multiple steps, each part follows standard techniques without requiring novel insight or complex problem-solving. |
| Spec | 1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{d\theta} = 4\sinh\alpha\) and \(\frac{dy}{d\theta} = 2\cosh\alpha\) so \(\frac{dy}{dx} = \frac{2\cosh\alpha}{4\sinh\alpha}\) | M1A1 | M1: Differentiates \(x\) and \(y\) and divides correctly. A1: Correct derivative in terms of \(\alpha\) |
| OR \(\frac{2x}{16} - \frac{2yy'}{4} = 0 \Rightarrow y' = \frac{x}{4y} = \frac{4\cosh\alpha}{8\sinh\alpha}\) | M1A1 | M1: Differentiates implicitly to obtain \(px - qyy' = 0\) and makes \(y'\) the subject. A1: Correct derivative in terms of \(\alpha\) |
| OR \(y = \frac{\sqrt{x^2-16}}{2} \Rightarrow y' = \frac{x}{2\sqrt{x^2-16}} = \frac{4\cosh\alpha}{2\sqrt{16\cosh^2\alpha - 16}}\) | M1A1 | M1: Differentiates explicitly to obtain \(y' = \frac{kx}{\sqrt{x^2-16}}\). A1: Correct derivative in terms of \(\alpha\) |
| Equation of tangent is \((y - 2\sinh\alpha) = \frac{\cosh\alpha}{2\sinh\alpha}(x - 4\cosh\alpha)\) (I) | M1 | Correct straight line method using their gradient in terms of \(\alpha\) |
| \(2y\sinh\alpha - 4\sinh^2\alpha = x\cosh\alpha - 4\cosh^2\alpha\) (II) | ||
| \(2y\sinh\alpha + 4(\cosh^2\alpha - \sinh^2\alpha) - x\cosh\alpha = 0 \Rightarrow 2y\sinh\alpha - x\cosh\alpha + 4 = 0\)* | A1* | See use of \(\cosh^2\alpha - \sinh^2\alpha = 1\) to give printed answer. (I) to * is A0, (II) to * is A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Puts \(x=0\) to give \(A\) is \(\left(0, \frac{-2}{\sinh\alpha}\right)\) | M1A1 | M1: Uses \(x=0\) in the given equation to find \(y\). A1: \(y = \frac{-2}{\sinh\alpha}\) or \(y = \frac{-4}{2\sinh\alpha}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2 = a^2(e^2-1) \Rightarrow a^2e^2 = 20\) | M1 | Uses the correct eccentricity formula to obtain a value for \(a^2e^2\) or \(ae\), or finds a value for \(e\) and multiplies by \(a\), or finds a value for \(e^2\) and multiplies by \(a^2\) |
| \(ae = \sqrt{20}\) or \(2\sqrt{5}\) | A1 | Correct value for \(ae\). Allow correct answer only |
| Gradient \(AS = \dfrac{\frac{2}{\sinh\alpha}}{2\sqrt{5}}\) or Gradient \(BS = -\dfrac{10\sinh\alpha}{2\sqrt{5}}\) | B1 | At least one correct gradient or vector (allow as "coordinates") in terms of \(\sinh\alpha\) (allow if also in terms of \(a\) and or \(e\)) |
| \(\dfrac{\frac{2}{\sinh\alpha}}{2\sqrt{5}} \times -\dfrac{10\sinh\alpha}{2\sqrt{5}} = -1\), so \(AS\) and \(BS\) are perpendicular | M1A1 | M1: Multiplies their AS and BS gradients or uses scalar product e.g. \(\overrightarrow{SB}\cdot\overrightarrow{SA}\) in terms of \(\sinh\alpha\) only and must be seen explicitly. A1: Product \(= -1\) or scalar product \(= 0\) with no errors and conclusion |
# Question 6:
## Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{d\theta} = 4\sinh\alpha$ and $\frac{dy}{d\theta} = 2\cosh\alpha$ so $\frac{dy}{dx} = \frac{2\cosh\alpha}{4\sinh\alpha}$ | M1A1 | M1: Differentiates $x$ and $y$ and divides correctly. A1: Correct derivative in terms of $\alpha$ |
| **OR** $\frac{2x}{16} - \frac{2yy'}{4} = 0 \Rightarrow y' = \frac{x}{4y} = \frac{4\cosh\alpha}{8\sinh\alpha}$ | M1A1 | M1: Differentiates implicitly to obtain $px - qyy' = 0$ and makes $y'$ the subject. A1: Correct derivative in terms of $\alpha$ |
| **OR** $y = \frac{\sqrt{x^2-16}}{2} \Rightarrow y' = \frac{x}{2\sqrt{x^2-16}} = \frac{4\cosh\alpha}{2\sqrt{16\cosh^2\alpha - 16}}$ | M1A1 | M1: Differentiates explicitly to obtain $y' = \frac{kx}{\sqrt{x^2-16}}$. A1: Correct derivative in terms of $\alpha$ |
| Equation of tangent is $(y - 2\sinh\alpha) = \frac{\cosh\alpha}{2\sinh\alpha}(x - 4\cosh\alpha)$ **(I)** | M1 | Correct straight line method using their gradient in terms of $\alpha$ |
| $2y\sinh\alpha - 4\sinh^2\alpha = x\cosh\alpha - 4\cosh^2\alpha$ **(II)** | | |
| $2y\sinh\alpha + 4(\cosh^2\alpha - \sinh^2\alpha) - x\cosh\alpha = 0 \Rightarrow 2y\sinh\alpha - x\cosh\alpha + 4 = 0$* | A1* | See use of $\cosh^2\alpha - \sinh^2\alpha = 1$ to give printed answer. **(I) to * is A0, (II) to * is A1** |
## Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Puts $x=0$ to give $A$ is $\left(0, \frac{-2}{\sinh\alpha}\right)$ | M1A1 | M1: Uses $x=0$ in the given equation to find $y$. A1: $y = \frac{-2}{\sinh\alpha}$ or $y = \frac{-4}{2\sinh\alpha}$ |
## Part 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2 = a^2(e^2-1) \Rightarrow a^2e^2 = 20$ | M1 | Uses the **correct** eccentricity formula to obtain a value for $a^2e^2$ or $ae$, or finds a value for $e$ and multiplies by $a$, or finds a value for $e^2$ and multiplies by $a^2$ |
| $ae = \sqrt{20}$ or $2\sqrt{5}$ | A1 | Correct value for $ae$. **Allow correct answer only** |
| Gradient $AS = \dfrac{\frac{2}{\sinh\alpha}}{2\sqrt{5}}$ **or** Gradient $BS = -\dfrac{10\sinh\alpha}{2\sqrt{5}}$ | B1 | At least one correct gradient or vector (allow as "coordinates") in terms of $\sinh\alpha$ (allow if also in terms of $a$ and or $e$) |
| $\dfrac{\frac{2}{\sinh\alpha}}{2\sqrt{5}} \times -\dfrac{10\sinh\alpha}{2\sqrt{5}} = -1$, so $AS$ and $BS$ are perpendicular | M1A1 | M1: Multiplies their AS and BS gradients or uses scalar product e.g. $\overrightarrow{SB}\cdot\overrightarrow{SA}$ in terms of $\sinh\alpha$ only and must be seen explicitly. A1: Product $= -1$ or scalar product $= 0$ with no errors and conclusion |
---6. The hyperbola $H$ has equation
$$\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 1$$
The line $l$ is a tangent to $H$ at the point $P ( 4 \cosh \alpha , 2 \sinh \alpha )$, where $\alpha$ is a constant, $\alpha \neq 0$
\begin{enumerate}[label=(\alph*)]
\item Using calculus, show that an equation for $l$ is
$$2 y \sinh \alpha - x \cosh \alpha + 4 = 0$$
The line $l$ cuts the $y$-axis at the point $A$.
\item Find the coordinates of $A$ in terms of $\alpha$.
The point $B$ has coordinates ( $0,10 \sinh \alpha$ ) and the point $S$ is the focus of $H$ for which $x > 0$
\item Show that the line segment $A S$ is perpendicular to the line segment $B S$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2014 Q6 [11]}}