| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola tangent intersection |
| Difficulty | Challenging +1.2 This is a standard Further Maths FP1 question on rectangular hyperbolas using parametric forms. Part (a) is routine calculus requiring implicit differentiation and substitution. Part (b) involves solving simultaneous equations with the tangent formula, which is a well-practiced technique in FP1. While it requires careful algebra and working with parameters, it follows a predictable pattern without requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = c^x x^{-1} \Rightarrow \frac{dy}{dx} = -c^x x^{-2}\) or (implicitly) \(y + x\frac{dy}{dx} = 0\) or (chain rule) \(\frac{dy}{dx} = -ct^{-2} \times \frac{1}{c}\) | M1 | \(\frac{dy}{dx} = \pm k x^{-2}\) or \(y + x\frac{dy}{dx} = 0\) or their \(\frac{d*}{d*}\) their \(\frac{d*}{d*}\) |
| When \(x = ct\), \(m_T = \frac{dy}{dx} = \frac{-c^2}{(ct)^2} = -\frac{1}{t^2}\) or at \(P\left(ct, \frac{c}{t}\right)\), \(m_T = \frac{dy}{dx} = -\frac{y}{x} = \frac{ct^{-1}}{ct} = -\frac{1}{t^2}\) | A1 | \(\frac{dy}{dx} = -\frac{1}{t^2}\) |
| T: \(y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)\) | M1 | Applies \(y - \frac{c}{t} = (\text{their } m_T)(x - ct)\) where their \(m_T\) has come from calculus |
| T: \(t^2y - ct = -x + ct\) | At least one line of working. | |
| T: \(t^2y + x = 2ct\) * | A1 cso * | Correct solution. |
| Answer | Marks | Guidance |
|---|---|---|
| \(t^2\left(\frac{3c}{5}\right) + \left(-\frac{8c}{5}\right) = 2ct\) | M1 | Substitutes \(\left(-\frac{8c}{5}, \frac{3c}{5}\right)\) into tangent. |
| \(3t^2 - 8 = 10t\) | A1 | Correct 3TQ in terms of \(t\). Can include uncancelled \(c\). |
| \(\{3t^2 - 10t - 8 = 0\} \Rightarrow (t-4)(3t+2) = 0 \Rightarrow t = ...\) | M1 | Attempt to solve their 3TQ for \(t\) |
| \(t = 4, \frac{2}{3} \Rightarrow A\left(4c, \frac{c}{4}\right), B\left(-\frac{3c}{5}, -\frac{2}{3}c\right)\) | M1 | Uses one of their values of \(t\) to find \(A\) or \(B\) |
| A1 | Correct coordinates. Condone \(A\) and \(B\) swapped or missing. |
| Answer | Marks | Guidance |
|---|---|---|
| \(y - \frac{3c}{5} = -\frac{1}{t^2}\left(x - \frac{8c}{5}\right)\) | M1 | Substitutes \(\left(ct, \frac{c}{t}\right)\) into \(y - \frac{3c}{5} = -\frac{1}{t^2}\left(x - \frac{8c}{5}\right)\) |
| \(\Rightarrow \frac{c}{t} - \frac{3c}{5} = -\frac{1}{t^2}\left(ct + \frac{8c}{5}\right)\) | ||
| \(3t^2 - 10t - 8\) | A1 | Correct 3TQ in terms of \(t\). Can include uncancelled \(c\). |
| then apply the original mark scheme. |
| Answer | Marks | Guidance |
|---|---|---|
| \(A\left(ct_1, \frac{c}{t_1}\right), B\left(ct_2, \frac{c}{t_2}\right)\) | M1 | Substitutes \(A\) and \(B\) into the equation of the tangent, solves for \(x\) and \(y\) |
| \(t_1^2y + x = 2ct_1\) | ||
| \(t_2^2y + x = 2ct_2\) | ||
| \(t_1 + t_2 = \frac{10}{3}, t_1 t_2 = -\frac{8}{3}\) | ||
| \(3t^2 - 8 = 10t\) | A1 | Correct 3TQ in terms of \(t_1\) or \(t_2\). Can include uncancelled \(c\). |
| then apply original scheme |
**Part (a)**
| $y = c^x x^{-1} \Rightarrow \frac{dy}{dx} = -c^x x^{-2}$ or (implicitly) $y + x\frac{dy}{dx} = 0$ or (chain rule) $\frac{dy}{dx} = -ct^{-2} \times \frac{1}{c}$ | M1 | $\frac{dy}{dx} = \pm k x^{-2}$ or $y + x\frac{dy}{dx} = 0$ or their $\frac{d*}{d*}$ their $\frac{d*}{d*}$ |
| When $x = ct$, $m_T = \frac{dy}{dx} = \frac{-c^2}{(ct)^2} = -\frac{1}{t^2}$ or at $P\left(ct, \frac{c}{t}\right)$, $m_T = \frac{dy}{dx} = -\frac{y}{x} = \frac{ct^{-1}}{ct} = -\frac{1}{t^2}$ | A1 | $\frac{dy}{dx} = -\frac{1}{t^2}$ |
| T: $y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)$ | M1 | Applies $y - \frac{c}{t} = (\text{their } m_T)(x - ct)$ where their $m_T$ has come from calculus |
| T: $t^2y - ct = -x + ct$ | | At least one line of working. |
| T: $t^2y + x = 2ct$ * | A1 cso * | Correct solution. |
**Part (b)**
| $t^2\left(\frac{3c}{5}\right) + \left(-\frac{8c}{5}\right) = 2ct$ | M1 | Substitutes $\left(-\frac{8c}{5}, \frac{3c}{5}\right)$ into tangent. |
| $3t^2 - 8 = 10t$ | A1 | Correct 3TQ in terms of $t$. Can include uncancelled $c$. |
| $\{3t^2 - 10t - 8 = 0\} \Rightarrow (t-4)(3t+2) = 0 \Rightarrow t = ...$ | M1 | Attempt to solve their 3TQ for $t$ |
| $t = 4, \frac{2}{3} \Rightarrow A\left(4c, \frac{c}{4}\right), B\left(-\frac{3c}{5}, -\frac{2}{3}c\right)$ | M1 | Uses one of their values of $t$ to find $A$ or $B$ |
| | A1 | Correct coordinates. Condone $A$ and $B$ swapped or missing. |
**ALT 1 (b)**
| $y - \frac{3c}{5} = -\frac{1}{t^2}\left(x - \frac{8c}{5}\right)$ | M1 | Substitutes $\left(ct, \frac{c}{t}\right)$ into $y - \frac{3c}{5} = -\frac{1}{t^2}\left(x - \frac{8c}{5}\right)$ |
| $\Rightarrow \frac{c}{t} - \frac{3c}{5} = -\frac{1}{t^2}\left(ct + \frac{8c}{5}\right)$ | | |
| $3t^2 - 10t - 8$ | A1 | Correct 3TQ in terms of $t$. Can include uncancelled $c$. |
| then apply the original mark scheme. | | |
**ALT 2 (b)**
| $A\left(ct_1, \frac{c}{t_1}\right), B\left(ct_2, \frac{c}{t_2}\right)$ | M1 | Substitutes $A$ and $B$ into the equation of the tangent, solves for $x$ and $y$ |
| $t_1^2y + x = 2ct_1$ | | |
| $t_2^2y + x = 2ct_2$ | | |
| $t_1 + t_2 = \frac{10}{3}, t_1 t_2 = -\frac{8}{3}$ | | |
| $3t^2 - 8 = 10t$ | A1 | Correct 3TQ in terms of $t_1$ or $t_2$. Can include uncancelled $c$. |
| then apply original scheme | | |
**Total: 9 marks**
---\begin{enumerate}
\item The rectangular hyperbola $H$ has equation $x y = c ^ { 2 }$, where $c$ is a positive constant.
\end{enumerate}
Given that $P \left( c t , \frac { c } { t } \right) , t \neq 0$, is a general point on $H$,\\
(a) use calculus to show that the equation of the tangent to $H$ at $P$ can be written as
$$t ^ { 2 } y + x = 2 c t$$
The points $A$ and $B$ lie on $H$.\\
The tangent to $H$ at $A$ and the tangent to $H$ at $B$ meet at the point $\left( - \frac { 8 c } { 5 } , \frac { 3 c } { 5 } \right)$.\\
Given that the $x$ coordinate of $A$ is positive,\\
(b) find, in terms of $c$, the coordinates of $A$ and the coordinates of $B$.\\
\hfill \mbox{\textit{Edexcel FP1 2018 Q5 [9]}}