| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola tangent equation |
| Difficulty | Standard +0.3 This is a standard FP1 rectangular hyperbola question requiring implicit differentiation to find the tangent equation, then finding x-intercepts and calculating a triangle area. While it involves parametric coordinates and multiple steps, each technique is routine for Further Maths students with no novel insight required—slightly easier than average A-level difficulty when considering the full spectrum. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{c^2}{x} = c^2x^{-1} \Rightarrow \frac{dy}{dx} = -c^2x^{-2} = -\frac{c^2}{x^2}\) | M1 | \(\frac{dy}{dx} = kx^{-2}\) OR correct use of product rule giving sum of two terms, one correct, and rhs = 0 |
| \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = -\frac{c}{t^2} \cdot \frac{1}{c}\) | their \(\frac{dy}{dt} \times \frac{1}{\text{their } \frac{dx}{dt}}\) | |
| \(\frac{dy}{dx} = -c^2x^{-2}\) or \(x\frac{dy}{dx} + y = 0\) or \(\frac{dy}{dx} = \frac{-c}{t^2} \cdot \frac{1}{c}\) or equivalent | A1 | Correct differentiation |
| \(y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)\) \((\times t^2)\) | dM1 | \(y - \frac{c}{t} = \) their \(m_T(x - ct)\) or \(y = mx + c\) with their \(m_T\) and \((ct, \frac{c}{t})\); \(m_T\) must come from calculus and be a function of \(t\) or \(c\) or both |
| \(t^2y + x = 2ct\) (allow \(x + t^2y = 2ct\)) | A1* | Correct solution only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 0 \Rightarrow x = \frac{ct^4 - c}{t^3} \Rightarrow A\left(\frac{ct^4-c}{t^3}, 0\right)\) | B1 | \(\frac{ct^4-c}{t^3}\) or equivalent form |
| \(y = 0 \Rightarrow x = 2ct \Rightarrow B(2ct, 0)\) | B1 | \(2ct\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB = \) "\(2ct\)" \(-\) "\(\frac{ct^4-c}{t^3}\)" and \(PA = ct^{-3}\sqrt{t^4+1}\), \(PB = ct^{-1}\sqrt{t^4+1}\) | M1 | Attempt to subtract their \(x\)-coordinates either way around |
| Area \(APB = \frac{1}{2} \times \text{their } AB \times \frac{c}{t}\) | M1 | Valid complete method for area of triangle in terms of \(t\) or \(c\) and \(t\) |
| \(= \frac{1}{2}\left(2ct - \frac{ct^4-c}{t^3}\right)\frac{c}{t} = \frac{c^2(t^4+1)}{2t^4}\) | ||
| \(= 8\left(1 + \frac{1}{t^4}\right)\) or \(\frac{8(t^4+1)}{t^4}\) or \(\frac{8t^4+8}{t^4}\) or \(8 + \frac{8}{t^4}\) | A1 | Use \(c=4\) and complete to one of given forms; final answer must be positive |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{c^2}{x} = c^2x^{-1} \Rightarrow \frac{dy}{dx} = -c^2x^{-2} = -\frac{c^2}{x^2}$ | M1 | $\frac{dy}{dx} = kx^{-2}$ OR correct use of product rule giving sum of two terms, one correct, and rhs = 0 |
| $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = -\frac{c}{t^2} \cdot \frac{1}{c}$ | | their $\frac{dy}{dt} \times \frac{1}{\text{their } \frac{dx}{dt}}$ |
| $\frac{dy}{dx} = -c^2x^{-2}$ or $x\frac{dy}{dx} + y = 0$ or $\frac{dy}{dx} = \frac{-c}{t^2} \cdot \frac{1}{c}$ or equivalent | A1 | Correct differentiation |
| $y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)$ $(\times t^2)$ | dM1 | $y - \frac{c}{t} = $ their $m_T(x - ct)$ or $y = mx + c$ with their $m_T$ and $(ct, \frac{c}{t})$; **$m_T$ must come from calculus and be a function of $t$ or $c$ or both** |
| $t^2y + x = 2ct$ (allow $x + t^2y = 2ct$) | A1* | Correct solution only |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 0 \Rightarrow x = \frac{ct^4 - c}{t^3} \Rightarrow A\left(\frac{ct^4-c}{t^3}, 0\right)$ | B1 | $\frac{ct^4-c}{t^3}$ or equivalent form |
| $y = 0 \Rightarrow x = 2ct \Rightarrow B(2ct, 0)$ | B1 | $2ct$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB = $ "$2ct$" $-$ "$\frac{ct^4-c}{t^3}$" and $PA = ct^{-3}\sqrt{t^4+1}$, $PB = ct^{-1}\sqrt{t^4+1}$ | M1 | Attempt to subtract their $x$-coordinates either way around |
| Area $APB = \frac{1}{2} \times \text{their } AB \times \frac{c}{t}$ | M1 | Valid complete method for area of triangle in terms of $t$ or $c$ and $t$ |
| $= \frac{1}{2}\left(2ct - \frac{ct^4-c}{t^3}\right)\frac{c}{t} = \frac{c^2(t^4+1)}{2t^4}$ | | |
| $= 8\left(1 + \frac{1}{t^4}\right)$ or $\frac{8(t^4+1)}{t^4}$ or $\frac{8t^4+8}{t^4}$ or $8 + \frac{8}{t^4}$ | A1 | Use $c=4$ and complete to one of given forms; final answer must be positive |
---6. The rectangular hyperbola $H$ has cartesian equation $x y = c ^ { 2 }$.
The point $P \left( c t , \frac { c } { t } \right) , t > 0$, is a general point on $H$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the tangent to $H$ at the point $P$ is
$$t ^ { 2 } y + x = 2 c t$$
An equation of the normal to $H$ at the point $P$ is $t ^ { 3 } x - t y = c t ^ { 4 } - c$
Given that the normal to $H$ at $P$ meets the $x$-axis at the point $A$ and the tangent to $H$ at $P$ meets the $x$-axis at the point $B$,
\item find, in terms of $c$ and $t$, the coordinates of $A$ and the coordinates of $B$.
Given that $c = 4$,
\item find, in terms of $t$, the area of the triangle $A P B$. Give your answer in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2014 Q6 [9]}}