| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Rectangular hyperbola normal re-intersection |
| Difficulty | Challenging +1.2 This is a Further Maths FP1 question requiring implicit differentiation to find the normal, then solving a cubic equation when the normal re-intersects the hyperbola. While it involves multiple steps and algebraic manipulation beyond standard A-level, the techniques are routine for FP1 students and the rectangular hyperbola is a standard topic with predictable methods. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(xy = 3\) or \(y = \frac{3}{x}\), differentiate to get \(x\frac{dy}{dx} + y = 0\) | M1A1 | Use of product rule: sum of two terms including \(dy/dx\), one correct; or \(\frac{dy}{dx} = kx^{-2}\) |
| \(\frac{dy}{dx} = \frac{-y}{x}\) or \(\frac{dy}{dx} = -\frac{3}{x^2}\) | Correct derivative: \(-3x^{-2}\) or \(-\frac{y}{x}\) | |
| Gradient of normal is \(\frac{x}{y}\) or \(\frac{x^2}{3}\) | M1 | Use of perpendicular gradient rule \(m_N m_T = -1\) |
| \(y - 3 = \frac{1}{3}(x-1)\) | M1 | \(y-3 =\) their \(m_N(x-1)\); or \(y = mx+c\) using their \(m_N\) and \((1,3)\) to find \(c\) |
| \(y = \frac{1}{3}x + \frac{8}{3}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \(y = \frac{3}{x}\) into their normal | M1 | |
| \(\frac{9}{x} - x = 8\) | ||
| \(x^2 + 8x - 9 = 0\) | A1 | Correct 3-term quadratic |
| \((x+9)(x-1) = 0\) | M1 | Attempt to solve 3-term quadratic |
| \(x = -9,\quad y = -\frac{1}{3}\) | A1, A1 |
# Question 4:
## Part 4(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $xy = 3$ or $y = \frac{3}{x}$, differentiate to get $x\frac{dy}{dx} + y = 0$ | M1A1 | Use of product rule: sum of two terms including $dy/dx$, one correct; or $\frac{dy}{dx} = kx^{-2}$ |
| $\frac{dy}{dx} = \frac{-y}{x}$ or $\frac{dy}{dx} = -\frac{3}{x^2}$ | | Correct derivative: $-3x^{-2}$ or $-\frac{y}{x}$ |
| Gradient of normal is $\frac{x}{y}$ or $\frac{x^2}{3}$ | M1 | Use of perpendicular gradient rule $m_N m_T = -1$ |
| $y - 3 = \frac{1}{3}(x-1)$ | M1 | $y-3 =$ their $m_N(x-1)$; or $y = mx+c$ using their $m_N$ and $(1,3)$ to find $c$ |
| $y = \frac{1}{3}x + \frac{8}{3}$ | A1 | |
## Part 4(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $y = \frac{3}{x}$ into their normal | M1 | |
| $\frac{9}{x} - x = 8$ | | |
| $x^2 + 8x - 9 = 0$ | A1 | Correct 3-term quadratic |
| $(x+9)(x-1) = 0$ | M1 | Attempt to solve 3-term quadratic |
| $x = -9,\quad y = -\frac{1}{3}$ | A1, A1 | |
---4. The hyperbola $H$ has equation
$$x y = 3$$
The point $Q ( 1,3 )$ is on $H$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the normal to $H$ at $Q$ in the form $y = a x + b$, where $a$ and $b$ are constants.\\
(5)
The normal at $Q$ intersects $H$ again at the point $R$.
\item Find the coordinates of $R$.\\
(5)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q4 [10]}}