| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Hyperbola tangent and geometric properties |
| Difficulty | Challenging +1.2 This is a structured multi-part question on hyperbolas using parametric coordinates. Parts (a)-(b) are routine (asymptotes and tangent derivation using implicit differentiation). Parts (c)-(d) require algebraic manipulation to find intersection points and calculate area, but follow a clear guided path with standard techniques. More challenging than average due to the hyperbolic function context and multi-step geometry, but the structure provides significant scaffolding. |
| Spec | 1.07s Parametric and implicit differentiation4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x,\ y = -x\) | B1 | Both required. Accept \(y = \pm x\) and \(x = \pm y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{\cosh t}{\sinh t}\) | B1 | Correct gradient |
| \(y - \sinh t = \frac{\cosh t}{\sinh t}(x - \cosh t)\) | M1 | Correct straight line method. For \(y = mx+c\) method, \(c\) must be found |
| \(y\sinh t = x\cosh t - (\cosh^2 t - \sinh^2 t)\) | ||
| \(y\sinh t = x\cosh t - 1\)* | A1* cso | Obtains printed answer with at least one intermediate step |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=x \Rightarrow x = \frac{1}{\cosh t - \sinh t},\ y = \frac{1}{\cosh t - \sinh t}\) | B1 | All four values correct. May be in exponential form e.g. \((e^t, e^t)\) and \((e^{-t}, -e^{-t})\) |
| \(y=-x \Rightarrow x = \frac{1}{\cosh t + \sinh t},\ y = \frac{-1}{\cosh t + \sinh t}\) | ||
| \(X = \frac{1}{2}\left(\frac{1}{\cosh t - \sinh t} + \frac{1}{\cosh t + \sinh t}\right)\) | M1 | Correct attempt at \(X\) or \(Y\). May be in exponential form e.g. \(\left(\frac{e^t+e^{-t}}{2}, \frac{e^t-e^{-t}}{2}\right)\) |
| \(X = \frac{1}{2}\left(\frac{\cosh t + \sinh t + \cosh t - \sinh t}{\cosh^2 t - \sinh^2 t}\right) = \cosh t\) | A1cso | Obtains \(X = \cosh t\) and \(Y = \sinh t\). May be shown using exponentials. |
| \(Y = \frac{1}{2}\left(\frac{\cosh t + \sinh t - \cosh t + \sinh t}{\cosh^2 t - \sinh^2 t}\right) = \sinh t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A = \frac{1}{2}\sqrt{\frac{2}{(\cosh t - \sinh t)^2}}\cdot\sqrt{\frac{2}{(\cosh t + \sinh t)^2}}\) or e.g. \(\frac{1}{2}\sqrt{2e^{2t}}\sqrt{2e^{-2t}}\) | M1 | Correct triangle area method |
| \(= \frac{1}{\cosh^2 t - \sinh^2 t} = 1\) | A1 | Obtains an area of 1 |
| So area is independent of \(t\) | A1ft | Concludes independence of \(t\) having obtained a constant area. Conclusion must include the word "independent" (or "not dependent") but not e.g. just QED |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x,\ y = -x$ | B1 | Both required. Accept $y = \pm x$ and $x = \pm y$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{\cosh t}{\sinh t}$ | B1 | Correct gradient |
| $y - \sinh t = \frac{\cosh t}{\sinh t}(x - \cosh t)$ | M1 | Correct straight line method. For $y = mx+c$ method, $c$ must be found |
| $y\sinh t = x\cosh t - (\cosh^2 t - \sinh^2 t)$ | | |
| $y\sinh t = x\cosh t - 1$* | A1* cso | Obtains printed answer with at least one intermediate step |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=x \Rightarrow x = \frac{1}{\cosh t - \sinh t},\ y = \frac{1}{\cosh t - \sinh t}$ | B1 | All four values correct. May be in exponential form e.g. $(e^t, e^t)$ and $(e^{-t}, -e^{-t})$ |
| $y=-x \Rightarrow x = \frac{1}{\cosh t + \sinh t},\ y = \frac{-1}{\cosh t + \sinh t}$ | | |
| $X = \frac{1}{2}\left(\frac{1}{\cosh t - \sinh t} + \frac{1}{\cosh t + \sinh t}\right)$ | M1 | Correct attempt at $X$ or $Y$. May be in exponential form e.g. $\left(\frac{e^t+e^{-t}}{2}, \frac{e^t-e^{-t}}{2}\right)$ |
| $X = \frac{1}{2}\left(\frac{\cosh t + \sinh t + \cosh t - \sinh t}{\cosh^2 t - \sinh^2 t}\right) = \cosh t$ | A1cso | Obtains $X = \cosh t$ and $Y = \sinh t$. May be shown using exponentials. |
| $Y = \frac{1}{2}\left(\frac{\cosh t + \sinh t - \cosh t + \sinh t}{\cosh^2 t - \sinh^2 t}\right) = \sinh t$ | | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A = \frac{1}{2}\sqrt{\frac{2}{(\cosh t - \sinh t)^2}}\cdot\sqrt{\frac{2}{(\cosh t + \sinh t)^2}}$ or e.g. $\frac{1}{2}\sqrt{2e^{2t}}\sqrt{2e^{-2t}}$ | M1 | Correct triangle area method |
| $= \frac{1}{\cosh^2 t - \sinh^2 t} = 1$ | A1 | Obtains an area of 1 |
| So area is independent of $t$ | A1ft | Concludes independence of $t$ having obtained a constant area. Conclusion must include the word "independent" (or "not dependent") but not e.g. just QED |
---\begin{enumerate}
\item The hyperbola $H$ is given by the equation $x ^ { 2 } - y ^ { 2 } = 1$\\
(a) Write down the equations of the two asymptotes of $H$.\\
(b) Show that an equation of the tangent to $H$ at the point $P ( \cosh t , \sinh t )$ is
\end{enumerate}
$$y \sinh t = x \cosh t - 1$$
The tangent at $P$ meets the asymptotes of $H$ at the points $Q$ and $R$.\\
(c) Show that $P$ is the midpoint of $Q R$.\\
(d) Show that the area of the triangle $O Q R$, where $O$ is the origin, is independent of $t$.\\
\hfill \mbox{\textit{Edexcel FP3 2015 Q6 [10]}}