| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola tangent intersection |
| Difficulty | Standard +0.8 This is a Further Maths FP1 question requiring parametric differentiation of a rectangular hyperbola, finding tangent equations, and solving simultaneous equations to find an intersection point. While the techniques are standard for FP1, the algebraic manipulation in part (c) to simplify the coordinates of R requires careful work with fractions and is more demanding than typical A-level questions. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y = 9x^{-1} \Rightarrow \frac{dy}{dx} = -9x^{-2}\) or \(xy=9 \Rightarrow x\frac{dy}{dx}+y=0\) or \(\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{-3}{p^2}\cdot\frac{1}{3}\) | M1 | \(\frac{dy}{dx} = kx^{-2}\); or correct product rule; or their \(\frac{dy}{dt}\times\frac{1}{\text{their}\frac{dx}{dt}}\) |
| \(\frac{dy}{dx} = -9x^{-2}\) or \(x\frac{dy}{dx}+y=0\) or \(\frac{dy}{dx} = \frac{-3}{p^2}\cdot\frac{1}{3}\) | A1 | Correct differentiation |
| \(y - \frac{3}{p} = -\frac{1}{p^2}(x-3p)\) | M1 | Applies \(y - \frac{3}{p} = (\text{their } m)(x-3p)\); their \(m\) must be a function of \(p\) and come from their \(dy/dx\) |
| \(x + p^2y = 6p\) | A1 | cso given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x + q^2y = 6q\) | B1 | Allow this to score here or in (c) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(6p - p^2y = 6q - q^2y\) | M1 | Attempt to obtain an equation in one variable \(x\) or \(y\) |
| \(y(q^2 - p^2) = 6(q-p) \Rightarrow y = \frac{6(q-p)}{q^2-p^2}\) and \(x(q^2-p^2) = 6pq(q-p) \Rightarrow x = \frac{6pq(q-p)}{q^2-p^2}\) | M1 | Attempt to isolate \(x\) or \(y\) — must reach \(x\) or \(y = f(p,q)\) or \(f(p)\) or \(f(q)\) |
| \(y = \frac{6}{p+q}\) | A1 | One correct simplified coordinate |
| \(x = \frac{6pq}{p+q}\) | A1 | Both coordinates correct and simplified |
## Question 9(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $y = 9x^{-1} \Rightarrow \frac{dy}{dx} = -9x^{-2}$ or $xy=9 \Rightarrow x\frac{dy}{dx}+y=0$ or $\frac{dy}{dx} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{-3}{p^2}\cdot\frac{1}{3}$ | M1 | $\frac{dy}{dx} = kx^{-2}$; or correct product rule; or their $\frac{dy}{dt}\times\frac{1}{\text{their}\frac{dx}{dt}}$ |
| $\frac{dy}{dx} = -9x^{-2}$ or $x\frac{dy}{dx}+y=0$ or $\frac{dy}{dx} = \frac{-3}{p^2}\cdot\frac{1}{3}$ | A1 | Correct differentiation |
| $y - \frac{3}{p} = -\frac{1}{p^2}(x-3p)$ | M1 | Applies $y - \frac{3}{p} = (\text{their } m)(x-3p)$; their $m$ must be a function of $p$ and come from their $dy/dx$ |
| $x + p^2y = 6p$ | A1 | cso **given answer** |
**Special case: if correct gradient is quoted could score M0A0M1A1. (4 marks)**
## Question 9(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $x + q^2y = 6q$ | B1 | Allow this to score here or in (c) |
**(1 mark)**
## Question 9(c):
| Working | Mark | Guidance |
|---------|------|----------|
| $6p - p^2y = 6q - q^2y$ | M1 | Attempt to obtain an equation in one variable $x$ or $y$ |
| $y(q^2 - p^2) = 6(q-p) \Rightarrow y = \frac{6(q-p)}{q^2-p^2}$ and $x(q^2-p^2) = 6pq(q-p) \Rightarrow x = \frac{6pq(q-p)}{q^2-p^2}$ | M1 | Attempt to isolate $x$ or $y$ — must reach $x$ or $y = f(p,q)$ or $f(p)$ or $f(q)$ |
| $y = \frac{6}{p+q}$ | A1 | One correct simplified coordinate |
| $x = \frac{6pq}{p+q}$ | A1 | Both coordinates correct and simplified |
**(4 marks) Total: 9**9. The rectangular hyperbola $H$ has cartesian equation $x y = 9$
The points $P \left( 3 p , \frac { 3 } { p } \right)$ and $Q \left( 3 q , \frac { 3 } { q } \right)$ lie on $H$, where $p \neq \pm q$.
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the tangent at $P$ is $x + p ^ { 2 } y = 6 p$.
\item Write down the equation of the tangent at $Q$.
The tangent at the point $P$ and the tangent at the point $Q$ intersect at $R$.
\item Find, as single fractions in their simplest form, the coordinates of $R$ in terms of $p$ and $q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2012 Q9 [9]}}