| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola normal re-intersection |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question on rectangular hyperbolas requiring parametric differentiation, normal equation derivation (with algebraic manipulation), and finding a second intersection point. While the techniques are standard FP1 content, part (b) requires careful algebraic work and part (c) involves solving a cubic equation after substituting the tangent into the hyperbola equation. The question demands more sophistication than typical A-level pure maths but is a standard FP1 exercise rather than requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(c^2 = 9\) | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \dfrac{9}{x} \Rightarrow \dfrac{dy}{dx} = -\dfrac{9}{x^2}\) | M1 | |
| Gradient of curve and tangent is \(-\dfrac{1}{t^2}\) | A1 | |
| Gradient of normal is \(-\dfrac{1}{\text{gradient of tangent}}\) | M1 | |
| Equation is \(y - \dfrac{3}{t} = t^2(x - 3t)\) giving printed answer | M1 A1 cso | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(t=2\), \(y = 4x + 1.5 - 24\) | M1 A1 | |
| \(\therefore \dfrac{9}{x} = 4x + 1.5 - 24\) | M1 | |
| Attempt to solve e.g. \(4x^2 - 22.5x - 9 = 0\) and formula | M1 A1 | |
| or factorise \(x = -\dfrac{3}{8}\); \(y = -24\) | M1 A1 | (7 marks) |
| (13 marks total) |
# Question 8:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $c^2 = 9$ | B1 | (1 mark) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \dfrac{9}{x} \Rightarrow \dfrac{dy}{dx} = -\dfrac{9}{x^2}$ | M1 | |
| Gradient of curve and tangent is $-\dfrac{1}{t^2}$ | A1 | |
| Gradient of normal is $-\dfrac{1}{\text{gradient of tangent}}$ | M1 | |
| Equation is $y - \dfrac{3}{t} = t^2(x - 3t)$ giving printed answer | M1 A1 cso | (5 marks) |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $t=2$, $y = 4x + 1.5 - 24$ | M1 A1 | |
| $\therefore \dfrac{9}{x} = 4x + 1.5 - 24$ | M1 | |
| Attempt to solve e.g. $4x^2 - 22.5x - 9 = 0$ and formula | M1 A1 | |
| or factorise $x = -\dfrac{3}{8}$; $y = -24$ | M1 A1 | (7 marks) |
| **(13 marks total)** | | |
---8. The rectangular hyperbola $H$ has equation $x y = c ^ { 2 }$.
The point ( $3 t , \frac { 3 } { t }$ ) is a general point on this hyperbola.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $c ^ { 2 }$.
\item Show that an equation of the normal to $H$ at the point ( $3 t , \frac { 3 } { t }$ ) is
$$y = t ^ { 2 } x + \left( \frac { 3 } { t } - 3 t ^ { 3 } \right)$$
The point $P$ on $H$ has coordinates (6, 1.5). The tangent to $H$ at $P$ meets the curve again at the point $Q$.
\item Find the coordinates of the point $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 Q8 [13]}}