| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola focus-directrix properties |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question combining parabola and hyperbola properties. Part (a) is routine recall of asymptote equations. Parts (b) require finding normals to a parabola with specified gradients (standard technique but requires careful algebra). Part (c) involves finding intersection points and computing a triangle area. While it requires multiple techniques and careful coordination across conic sections, each individual step follows standard FP1 procedures without requiring novel insight. The multi-step nature and Further Maths context place it moderately above average difficulty. |
| Spec | 1.07m Tangents and normals: gradient and equations4.03d Linear transformations 2D: reflection, rotation, enlargement, shear |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equations of asymptotes of \(H\) are \(y = \pm\frac{1}{2}x\), e.g. \(y = \pm\frac{3}{6}x\) or \(\frac{x}{6} = \pm\frac{y}{3}\) | B1 | Correct equations for asymptotes of \(H\), any form, need not be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For parabola \(2y\frac{dy}{dx} = 32 \Rightarrow \frac{dy}{dx} = ...\) or \(y = \sqrt{32x} \Rightarrow \frac{dy}{dx} = ...x^{-\frac{1}{2}}\) or \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = ...\) | M1 | A correct method to find the gradient of the parabola |
| Finds gradient of normal using \(m_N = \frac{-1}{\text{their } \frac{dy}{dx}}\); \(m_N = -\frac{y}{16}\) or \(-t\) or \(-\frac{\sqrt{x}}{2\sqrt{2}}\), so \(m_N = (\pm)\frac{1}{2} \Rightarrow y = (\pm)8, x = 2\) | M1 | Finds gradient of normal and sets normal gradient equal to asymptote gradient to obtain at least one point on \(C\) |
| Finds equation of either \(l_1\) or \(l_2\): \(y - \text{"8"} = \text{"their } m_N\text{"}(x - \text{"2"})\) or \(y - \text{"}-8\text{"} = \text{"their } m_N\text{"}(x - \text{"2"})\) | M1 | Finds the equation of either \(l_1\) or \(l_2\) |
| \(l_1\) is \(y + 8 = \frac{1}{2}(x-2)\) and \(l_2\) is \(y - 8 = -\frac{1}{2}(x-2)\); \(y = \frac{1}{2}x - 9\) and \(y = -\frac{1}{2}x + 9\) | A1 | Correct equation for each normal; ignore labelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Meet \(H \Rightarrow \frac{x^2}{36} - \frac{(\pm(\frac{1}{2}x-9))^2}{9} = 1 \Rightarrow \frac{x^2}{36} - \frac{\frac{1}{4}x^2 - 9x + 81}{9} = 1 \Rightarrow x = ...\) or \(\frac{(18\pm 2y)^2}{36} - \frac{y^2}{9} = 1 \Rightarrow \frac{81\pm 18y + y^2}{9} - \frac{y^2}{9} = 1 \Rightarrow y = ...\) | M1 | Substitutes for \(x\) or \(y\) into equation of hyperbola and solves |
| One correct point of intersection \((10, \pm4)\) | A1 | Achieves one correct coordinate \(x=10\) and \(y=\pm4\) |
| Area \(OPQ\) is \(\frac{1}{2} \times 10 \times (4-(-4)) = ...\) | dM1 | Dependent on previous method; correct method for area of triangle, e.g. \(\frac{1}{2} \times \text{their } 10 \times \text{twice their } 4\) or equivalent determinant methods |
| \(= 40\) | A1 | Correct answer only |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equations of asymptotes of $H$ are $y = \pm\frac{1}{2}x$, e.g. $y = \pm\frac{3}{6}x$ or $\frac{x}{6} = \pm\frac{y}{3}$ | B1 | Correct equations for asymptotes of $H$, any form, need not be simplified |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For parabola $2y\frac{dy}{dx} = 32 \Rightarrow \frac{dy}{dx} = ...$ or $y = \sqrt{32x} \Rightarrow \frac{dy}{dx} = ...x^{-\frac{1}{2}}$ or $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = ...$ | M1 | A correct method to find the gradient of the parabola |
| Finds gradient of normal using $m_N = \frac{-1}{\text{their } \frac{dy}{dx}}$; $m_N = -\frac{y}{16}$ or $-t$ or $-\frac{\sqrt{x}}{2\sqrt{2}}$, so $m_N = (\pm)\frac{1}{2} \Rightarrow y = (\pm)8, x = 2$ | M1 | Finds gradient of normal and sets normal gradient equal to asymptote gradient to obtain at least one point on $C$ |
| Finds equation of either $l_1$ or $l_2$: $y - \text{"8"} = \text{"their } m_N\text{"}(x - \text{"2"})$ or $y - \text{"}-8\text{"} = \text{"their } m_N\text{"}(x - \text{"2"})$ | M1 | Finds the equation of either $l_1$ or $l_2$ |
| $l_1$ is $y + 8 = \frac{1}{2}(x-2)$ and $l_2$ is $y - 8 = -\frac{1}{2}(x-2)$; $y = \frac{1}{2}x - 9$ and $y = -\frac{1}{2}x + 9$ | A1 | Correct equation for each normal; ignore labelling |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Meet $H \Rightarrow \frac{x^2}{36} - \frac{(\pm(\frac{1}{2}x-9))^2}{9} = 1 \Rightarrow \frac{x^2}{36} - \frac{\frac{1}{4}x^2 - 9x + 81}{9} = 1 \Rightarrow x = ...$ or $\frac{(18\pm 2y)^2}{36} - \frac{y^2}{9} = 1 \Rightarrow \frac{81\pm 18y + y^2}{9} - \frac{y^2}{9} = 1 \Rightarrow y = ...$ | M1 | Substitutes for $x$ or $y$ into equation of hyperbola and solves |
| One correct point of intersection $(10, \pm4)$ | A1 | Achieves one correct coordinate $x=10$ and $y=\pm4$ |
| Area $OPQ$ is $\frac{1}{2} \times 10 \times (4-(-4)) = ...$ | dM1 | Dependent on previous method; correct method for area of triangle, e.g. $\frac{1}{2} \times \text{their } 10 \times \text{twice their } 4$ or equivalent determinant methods |
| $= 40$ | A1 | Correct answer only |
---\begin{enumerate}
\item The parabola $C$ has equation
\end{enumerate}
$$y ^ { 2 } = 32 x$$
and the hyperbola $H$ has equation
$$\frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 9 } = 1$$
(a) Write down the equations of the asymptotes of $H$.
The line $l _ { 1 }$ is normal to $C$ and parallel to the asymptote of $H$ with positive gradient. The line $l _ { 2 }$ is normal to $C$ and parallel to the asymptote of $H$ with negative gradient.\\
(b) Determine\\
(i) an equation for $l _ { 1 }$\\
(ii) an equation for $l _ { 2 }$
The lines $l _ { 1 }$ and $l _ { 2 }$ meet $H$ at the points $P$ and $Q$ respectively.\\
(c) Find the area of the triangle $O P Q$, where $O$ is the origin.
\hfill \mbox{\textit{Edexcel FP1 2021 Q5 [9]}}