| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola normal re-intersection |
| Difficulty | Challenging +1.2 This is a standard Further Maths rectangular hyperbola question requiring implicit differentiation to find the gradient, then finding the normal equation (part given), and solving a simultaneous equation with the hyperbola. While it involves multiple steps and Further Maths content, the techniques are routine for FP1 students with no novel problem-solving required. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx}=-\frac{9}{x^2}\) or \(\frac{dy}{dx}=-\frac{y}{x}\) or \(\frac{dy}{dx}=-\frac{1}{t^2}\) | M1 | Differentiates to obtain \(\frac{k}{x^2}\) and substitutes \(x=6\); or uses implicit differentiation \(\frac{dy}{dx}=-\frac{y}{x}\) and substitutes \(x\) and \(y\); or uses parametric differentiation \(\frac{dy}{dx}=-\frac{1}{t^2}\) and substitutes \(t=2\) |
| Gradient at \(x=6\) or \(t=2\) is \(-\frac{9}{36}\) or \(-\frac{3}{6}\) or \(-\frac{1}{4}\) | A1 | Accept any equivalent i.e. \(-0.25\) etc |
| Gradient of normal is \(-\frac{1}{m}\ (=4)\) | M1 | Uses negative reciprocal of their gradient |
| Equation of normal: \(y-\frac{3}{2}=4(x-6)\) | dM1 | \(y-y_1=m(x-x_1)\) with \(\left(6,\frac{3}{2}\right)\) or \(y=mx+c\) and sub \(\left(6,\frac{3}{2}\right)\) to find \(c\) |
| \(2y-8x+45=0\) given answer | A1* (5) | cso: Correct answer with no errors seen in solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{18}{x}-8x+45=0\) or \(2y-\frac{72}{y}+45=0\) or \(x(4x-22.5)=9\) etc | M1 | Obtains equation in one variable, \(x\) or \(y\) |
| \(8x^2-45x-18=0\) or \(2y^2+45y-72=0\) | ||
| So \(x=-\frac{3}{8}\) or \(y=-24\) | A1 | Correct value of \(x\) or correct value of \(y\) |
| Finds other ordinate: \(\left(-\frac{3}{8}, -24\right)\) | M1 A1 | M1: Finds second coordinate using \(xy=9\) or solving second quadratic or equation of normal. A1: Correct coordinates that can be written as \(x=\ldots, y=\ldots\) |
| ALT: Sub \(\left(3t, \frac{3}{t}\right)\) in \(2y-8x+45=0 \Rightarrow t=-\frac{1}{8}\) | M1A1 | |
| Sub \(t=-\frac{1}{8}\) in \(\left(3t,\frac{3}{t}\right) \Rightarrow \left(-\frac{3}{8},-24\right)\) | M1A1 (4)(9 marks) |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=-\frac{9}{x^2}$ or $\frac{dy}{dx}=-\frac{y}{x}$ or $\frac{dy}{dx}=-\frac{1}{t^2}$ | M1 | Differentiates to obtain $\frac{k}{x^2}$ and substitutes $x=6$; or uses implicit differentiation $\frac{dy}{dx}=-\frac{y}{x}$ and substitutes $x$ and $y$; or uses parametric differentiation $\frac{dy}{dx}=-\frac{1}{t^2}$ and substitutes $t=2$ |
| Gradient at $x=6$ or $t=2$ is $-\frac{9}{36}$ or $-\frac{3}{6}$ or $-\frac{1}{4}$ | A1 | Accept any equivalent i.e. $-0.25$ etc |
| Gradient of normal is $-\frac{1}{m}\ (=4)$ | M1 | Uses negative reciprocal of their gradient |
| Equation of normal: $y-\frac{3}{2}=4(x-6)$ | dM1 | $y-y_1=m(x-x_1)$ with $\left(6,\frac{3}{2}\right)$ or $y=mx+c$ and sub $\left(6,\frac{3}{2}\right)$ to find $c$ |
| $2y-8x+45=0$ **given answer** | A1* (5) | cso: Correct answer with no errors seen in solution |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{18}{x}-8x+45=0$ or $2y-\frac{72}{y}+45=0$ or $x(4x-22.5)=9$ etc | M1 | Obtains equation in one variable, $x$ or $y$ |
| $8x^2-45x-18=0$ or $2y^2+45y-72=0$ | | |
| So $x=-\frac{3}{8}$ or $y=-24$ | A1 | Correct value of $x$ or correct value of $y$ |
| Finds other ordinate: $\left(-\frac{3}{8}, -24\right)$ | M1 A1 | M1: Finds second coordinate using $xy=9$ or solving second quadratic or equation of normal. A1: Correct coordinates that can be written as $x=\ldots, y=\ldots$ |
| **ALT:** Sub $\left(3t, \frac{3}{t}\right)$ in $2y-8x+45=0 \Rightarrow t=-\frac{1}{8}$ | M1A1 | |
| Sub $t=-\frac{1}{8}$ in $\left(3t,\frac{3}{t}\right) \Rightarrow \left(-\frac{3}{8},-24\right)$ | M1A1 (4)(9 marks) | |5. The rectangular hyperbola $H$ has equation $x y = 9$
The point $A$ on $H$ has coordinates $\left( 6 , \frac { 3 } { 2 } \right)$.
\begin{enumerate}[label=(\alph*)]
\item Show that the normal to $H$ at the point $A$ has equation
$$2 y - 8 x + 45 = 0$$
The normal at $A$ meets $H$ again at the point $B$.
\item Find the coordinates of $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2015 Q5 [9]}}