| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola locus problems |
| Difficulty | Challenging +1.2 This is a multi-step Further Maths question requiring implicit differentiation, parametric tangent equations, perpendicular line geometry, and locus derivation. Part (a) is guided calculus application, while part (b) requires algebraic manipulation to eliminate the parameter. However, the techniques are standard for FP3 hyperbola questions with clear scaffolding, making it moderately above average difficulty but not requiring exceptional insight. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dt} = 4\sec t\tan t\), \(\quad\frac{dy}{dt} = 2\sec^2 t\) | B1 (both) | Both derivatives correct |
| \(\frac{dy}{dx} = \frac{2\sec^2 t}{4\sec t\tan t} = \frac{1}{2\sin t}\) | M1 | Dividing derivatives |
| \(y - 2\tan t = \frac{1}{2\sin t}(x - 4\sec t)\) | M1 A1 | Tangent line equation |
| \(2y\sin t = x - 4\cos t\) ✱ | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient of \(l_2\) is \(-2\sin t\) | M1 | Perpendicular gradient |
| \(y = -2x\sin t\) | A1 | Line through origin |
| \(2(-2x\sin t)\sin t = x - 4\cos t \Rightarrow x = \frac{4\cos t}{1+4\sin^2 t}\) | M1 A1 | Solving simultaneously (1) |
| \(y = \frac{-8\sin t\cos t}{1+4\sin^2 t}\) | M1 A1 | Substituting back (2) |
| \((x^2+y^2)^2 = \left(\frac{16\cos^2 t}{(1+4\sin^2 t)^2} + \frac{64\sin^2 t\cos^2 t}{(1+4\sin^2 t)^2}\right)^2 = \frac{256\cos^4 t}{(1+4\sin^2 t)^2}\) | M1 | |
| \(16x^2 - 4y^2 = \frac{256\cos^2 t}{(1+4\sin^2 t)^2} - \frac{256\sin^2 t\cos^2 t}{(1+4\sin^2 t)^2} = \frac{256\cos^4 t}{(1+4\sin^2 t)^2}\) | A1 | (8 marks, 13 overall) |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 4\sec t\tan t$, $\quad\frac{dy}{dt} = 2\sec^2 t$ | B1 (both) | Both derivatives correct |
| $\frac{dy}{dx} = \frac{2\sec^2 t}{4\sec t\tan t} = \frac{1}{2\sin t}$ | M1 | Dividing derivatives |
| $y - 2\tan t = \frac{1}{2\sin t}(x - 4\sec t)$ | M1 A1 | Tangent line equation |
| $2y\sin t = x - 4\cos t$ ✱ | A1 | **(5 marks)** |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient of $l_2$ is $-2\sin t$ | M1 | Perpendicular gradient |
| $y = -2x\sin t$ | A1 | Line through origin |
| $2(-2x\sin t)\sin t = x - 4\cos t \Rightarrow x = \frac{4\cos t}{1+4\sin^2 t}$ | M1 A1 | Solving simultaneously (1) |
| $y = \frac{-8\sin t\cos t}{1+4\sin^2 t}$ | M1 A1 | Substituting back (2) |
| $(x^2+y^2)^2 = \left(\frac{16\cos^2 t}{(1+4\sin^2 t)^2} + \frac{64\sin^2 t\cos^2 t}{(1+4\sin^2 t)^2}\right)^2 = \frac{256\cos^4 t}{(1+4\sin^2 t)^2}$ | M1 | |
| $16x^2 - 4y^2 = \frac{256\cos^2 t}{(1+4\sin^2 t)^2} - \frac{256\sin^2 t\cos^2 t}{(1+4\sin^2 t)^2} = \frac{256\cos^4 t}{(1+4\sin^2 t)^2}$ | A1 | **(8 marks, 13 overall)** |8. The hyperbola $H$ has equation $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 1$.
The line $l _ { 1 }$ is the tangent to $H$ at the point $P ( 4 \sec t , 2 \tan t )$.
\begin{enumerate}[label=(\alph*)]
\item Use calculus to show that an equation of $l _ { 1 }$ is
$$2 y \sin t = x - 4 \cos t$$
The line $l _ { 2 }$ passes through the origin and is perpendicular to $l _ { 1 }$.\\
The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $Q$.
\item Show that, as $t$ varies, an equation of the locus of $Q$ is
$$\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 16 x ^ { 2 } - 4 y ^ { 2 }$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2010 Q8 [13]}}