| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola area calculations |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring implicit differentiation of a rectangular hyperbola, finding a tangent equation in parametric form, then calculating intercepts and using area constraints to find a constant. While the techniques are standard FP1 content, the parametric approach and algebraic manipulation across multiple steps elevates it above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | 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 product rule with sum of two terms, rhs \(=0\) |
| \(\frac{dy}{dx}=-c^2x^{-2}\) or equivalent | A1 | Correct differentiation |
| \(m_T = \frac{dy}{dx} = -\frac{1}{t^2}\) | — | \(-\frac{1}{t^2}\) |
| \(y-\frac{c}{t}=-\frac{1}{t^2}(x-ct)\) | M1 | Uses \(y-\frac{c}{t}=m_T(x-ct)\); \(m_T\) must come from calculus |
| \(x+t^2y=2ct\) | A1* | Correct solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y=0 \Rightarrow x=2ct \Rightarrow A(2ct,0)\) | B1 | \(x=2ct\) seen or implied |
| \(x=0 \Rightarrow y=\frac{2ct}{t^2}=\frac{2c}{t} \Rightarrow B\left(0,\frac{2c}{t}\right)\) | B1 | \(y=\frac{2ct}{t^2}\) or \(\frac{2c}{t}\) seen or implied |
| Area \(OAB=36 \Rightarrow \frac{1}{2}(2ct)\left(\frac{2c}{t}\right)=36\) | M1 | Applies \(\frac{1}{2}(\text{their } x)(\text{their } y)=36\) |
| \(\Rightarrow 2c^2=36 \Rightarrow c^2=18 \Rightarrow c=3\sqrt{2}\) | A1 | \(c=3\sqrt{2}\) (Do not allow \(c=\pm3\sqrt{2}\)) |
# Question 8:
## Part (a):
| Answer | 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 product rule with sum of two terms, rhs $=0$ |
| $\frac{dy}{dx}=-c^2x^{-2}$ or equivalent | A1 | Correct differentiation |
| $m_T = \frac{dy}{dx} = -\frac{1}{t^2}$ | — | $-\frac{1}{t^2}$ |
| $y-\frac{c}{t}=-\frac{1}{t^2}(x-ct)$ | M1 | Uses $y-\frac{c}{t}=m_T(x-ct)$; $m_T$ must come from calculus |
| $x+t^2y=2ct$ | A1* | Correct solution |
**Note:** Candidates who state $m_T=-\frac{1}{t^2}$ with no justification score **no marks** in (a).
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y=0 \Rightarrow x=2ct \Rightarrow A(2ct,0)$ | B1 | $x=2ct$ seen or implied |
| $x=0 \Rightarrow y=\frac{2ct}{t^2}=\frac{2c}{t} \Rightarrow B\left(0,\frac{2c}{t}\right)$ | B1 | $y=\frac{2ct}{t^2}$ or $\frac{2c}{t}$ seen or implied |
| Area $OAB=36 \Rightarrow \frac{1}{2}(2ct)\left(\frac{2c}{t}\right)=36$ | M1 | Applies $\frac{1}{2}(\text{their } x)(\text{their } y)=36$ |
| $\Rightarrow 2c^2=36 \Rightarrow c^2=18 \Rightarrow c=3\sqrt{2}$ | A1 | $c=3\sqrt{2}$ (**Do not allow** $c=\pm3\sqrt{2}$) |
---8. The rectangular hyperbola $H$ has equation $x y = c ^ { 2 }$, where $c$ is a positive constant.
The point $P \left( c t , \frac { c } { t } \right) , t \neq 0$, is a general point on $H$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation for the tangent to $H$ at $P$ is
$$x + t ^ { 2 } y = 2 c t$$
The tangent to $H$ at the point $P$ meets the $x$-axis at the point $A$ and the $y$-axis at the point $B$. Given that the area of the triangle $O A B$, where $O$ is the origin, is 36 ,
\item find the exact value of $c$, expressing your answer in the form $k \sqrt { } 2$, where $k$ is an integer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2012 Q8 [8]}}