| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola tangent intersection |
| Difficulty | Standard +0.8 This is a Further Maths question requiring implicit differentiation to find the tangent equation (part a), then solving a system involving two parametric tangent equations simultaneously (part b). While the techniques are standard for FP1, the algebraic manipulation with parameters and solving the resulting system elevates this above typical A-level questions. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y = \frac{36}{x} = 36x^{-1} \Rightarrow \frac{dy}{dx} = -36x^{-2} = -\frac{36}{x^2}\) | M1 | An attempt at \(\frac{dy}{dx}\), or \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) |
| At \(\left(6t, \frac{6}{t}\right)\): \(\frac{dy}{dx} = -\frac{36}{(6t)^2}\) | M1 | An attempt at \(\frac{dy}{dx}\) in terms of \(t\) |
| \(m_T = \frac{dy}{dx} = -\frac{1}{t^2}\) | A1 | \(\frac{dy}{dx} = -\frac{1}{t^2}\) Must see working to award here |
| \(y - \frac{6}{t} = -\frac{1}{t^2}(x - 6t)\) | M1 | Applies \(y - \frac{6}{t} =\) their \(m_T(x-6t)\) |
| \(T: y = -\frac{1}{t^2}x + \frac{12}{t}\) | A1 cso | Correct solution |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Both tangents meet at \((-9, 12)\): \(12 = -\frac{1}{t^2}(-9) + \frac{12}{t}\) | M1 | Substituting \((-9,12)\) into T |
| \(12t^2 = 9 + 12t \Rightarrow 12t^2 - 12t - 9 = 0 \Rightarrow 4t^2 - 4t - 3 = 0\) | M1 | An attempt to form a "3 term quadratic" |
| \((2t-3)(2t+1) = 0\) | M1 | An attempt to factorise |
| \(t = \frac{3}{2}, -\frac{1}{2}\) | A1 | \(t = \frac{3}{2}, -\frac{1}{2}\) |
| \(t = \frac{3}{2} \Rightarrow x = 9,\ y = 4 \Rightarrow (9,4)\) | M1 | An attempt to substitute either \(t\) value into \(x\) and \(y\) |
| At least one of \((9,4)\) or \((-3,-12)\) | A1 | At least one correct point |
| \(t = -\frac{1}{2} \Rightarrow x = -3,\ y = -12 \Rightarrow (-3,-12)\) | A1 | Both \((9,4)\) and \((-3,-12)\) |
| Subtotal: 7 marks | Total: 12 marks |
## Question 10:
$xy = 36$ at $\left(6t, \frac{6}{t}\right)$
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $y = \frac{36}{x} = 36x^{-1} \Rightarrow \frac{dy}{dx} = -36x^{-2} = -\frac{36}{x^2}$ | M1 | An attempt at $\frac{dy}{dx}$, or $\frac{dy}{dt}$ and $\frac{dx}{dt}$ |
| At $\left(6t, \frac{6}{t}\right)$: $\frac{dy}{dx} = -\frac{36}{(6t)^2}$ | M1 | An attempt at $\frac{dy}{dx}$ in terms of $t$ |
| $m_T = \frac{dy}{dx} = -\frac{1}{t^2}$ | A1 | $\frac{dy}{dx} = -\frac{1}{t^2}$ **Must see working to award here** |
| $y - \frac{6}{t} = -\frac{1}{t^2}(x - 6t)$ | M1 | Applies $y - \frac{6}{t} =$ their $m_T(x-6t)$ |
| $T: y = -\frac{1}{t^2}x + \frac{12}{t}$ | A1 cso | Correct solution |
**Subtotal: 5 marks**
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| Both tangents meet at $(-9, 12)$: $12 = -\frac{1}{t^2}(-9) + \frac{12}{t}$ | M1 | Substituting $(-9,12)$ into **T** |
| $12t^2 = 9 + 12t \Rightarrow 12t^2 - 12t - 9 = 0 \Rightarrow 4t^2 - 4t - 3 = 0$ | M1 | An attempt to form a "3 term quadratic" |
| $(2t-3)(2t+1) = 0$ | M1 | An attempt to factorise |
| $t = \frac{3}{2}, -\frac{1}{2}$ | A1 | $t = \frac{3}{2}, -\frac{1}{2}$ |
| $t = \frac{3}{2} \Rightarrow x = 9,\ y = 4 \Rightarrow (9,4)$ | M1 | An attempt to substitute either $t$ value into $x$ and $y$ |
| At least one of $(9,4)$ or $(-3,-12)$ | A1 | At least one correct point |
| $t = -\frac{1}{2} \Rightarrow x = -3,\ y = -12 \Rightarrow (-3,-12)$ | A1 | Both $(9,4)$ and $(-3,-12)$ |
**Subtotal: 7 marks | Total: 12 marks**
---10. The point $P \left( 6 t , \frac { 6 } { t } \right) , t \neq 0$, lies on the rectangular hyperbola $H$ with equation $x y = 36$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation for the tangent to $H$ at $P$ is
$$y = - \frac { 1 } { t ^ { 2 } } x + \frac { 12 } { t }$$
The tangent to $H$ at the point $A$ and the tangent to $H$ at the point $B$ meet at the point $( - 9,12 )$.
\item Find the coordinates of $A$ and $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2011 Q10 [12]}}