| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| 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 to derive the tangent equation (part a), then solving a system involving two simultaneous equations with parameters to find intersection points (part b). The algebraic manipulation with parameters t₁ and t₂, including using Vieta's formulas or symmetric functions, elevates this above standard calculus questions. It's moderately challenging for Further Maths but not exceptionally difficult. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{c^2}{x}\), \(\frac{dy}{dx} = -c^2x^{-2}\) | B1 | |
| \(\frac{dy}{dx} = -\frac{c^2}{(ct)^2} = -\frac{1}{t^2}\) | M1 | Without \(x\) or \(y\) |
| \(y - \frac{c}{t} = -\frac{1}{t^2}(x - ct) \Rightarrow t^2y + x = 2ct\) | M1 A1cso | \((*)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \((15c, -c)\): \(-ct^2 + 15c = 2ct\) | M1 | |
| \(t^2 + 2t - 15 = 0\) | A1 | |
| \((t+5)(t-3) = 0 \Rightarrow t = -5,\ t = 3\) | M1 A1 | Correct absolute factors for M1; accept quadratic formula |
| Points are \(\left(-5c, -\frac{c}{5}\right)\) and \(\left(3c, \frac{c}{3}\right)\) | A1 | Both |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{c^2}{x}$, $\frac{dy}{dx} = -c^2x^{-2}$ | B1 | |
| $\frac{dy}{dx} = -\frac{c^2}{(ct)^2} = -\frac{1}{t^2}$ | M1 | Without $x$ or $y$ |
| $y - \frac{c}{t} = -\frac{1}{t^2}(x - ct) \Rightarrow t^2y + x = 2ct$ | M1 A1cso | $(*)$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $(15c, -c)$: $-ct^2 + 15c = 2ct$ | M1 | |
| $t^2 + 2t - 15 = 0$ | A1 | |
| $(t+5)(t-3) = 0 \Rightarrow t = -5,\ t = 3$ | M1 A1 | Correct absolute factors for M1; accept quadratic formula |
| Points are $\left(-5c, -\frac{c}{5}\right)$ and $\left(3c, \frac{c}{3}\right)$ | A1 | Both |
---7. The rectangular hyperbola $H$ has equation $x y = c ^ { 2 }$, where $c$ is a constant.
The point $P \left( c t , \frac { c } { t } \right)$ is a general point on $H$.
\begin{enumerate}[label=(\alph*)]
\item Show that the tangent to $H$ at $P$ has equation
$$t ^ { 2 } y + x = 2 c t$$
The tangents to $H$ at the points $A$ and $B$ meet at the point $( 15 c , - c )$.
\item Find, in terms of $c$, the coordinates of $A$ and $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2010 Q7 [9]}}