| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola locus problems |
| Difficulty | Challenging +1.3 This is a multi-part Further Maths question requiring implicit differentiation to find a tangent equation (part a), then finding the intersection of perpendicular lines and eliminating the parameter to derive a locus (part b). While it involves several steps and coordinate geometry manipulation, the techniques are standard for F3 level with clear scaffolding. The locus derivation requires algebraic stamina but follows a predictable method, making it moderately above average difficulty. |
| Spec | 1.07m Tangents and normals: gradient and equations4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{dx}{dt} = 4\sec t \tan t\), \(\quad \frac{dy}{dt} = 2\sec^2 t\) | B1 (both) | |
| \(\frac{dy}{dx} = \frac{2\sec^2 t}{4\sec t \tan t} = \frac{1}{2\sin t}\) | M1 | |
| \(y - 2\tan t = \frac{1}{2\sin t}(x - 4\sec t)\) | M1 A1 | |
| \(2y\sin t - \frac{4\sin^2 t}{\cos t} = x - \frac{4}{\cos t}\) | ||
| \(2y\sin t = x - \frac{4-4\sin^2 t}{\cos t} = x - 4\cos t\) ✱ | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Gradient of \(l_2\) is \(-2\sin t\) | M1 | |
| \(y = -2x\sin t\) | A1 | |
| \(2(-2x\sin t)\sin t = x - 4\cos t \Rightarrow x = \frac{4\cos t}{1+4\sin^2 t}\) | M1 A1 | (1) |
| \(y = \frac{-8\sin t \cos t}{1+4\sin^2 t}\) | M1 A1 | |
| \((x^2+y^2)^2 = \left(\frac{16\cos^2 t}{(1+4\sin^2 t)^2} + \frac{64\sin^2 t\cos^2 t}{(1+4\sin^2 t)^2}\right)^2\) | ||
| \(= \frac{256\cos^4 t(1+4\sin^2 t)^2}{(1+4\sin^2 t)^4} = \frac{256\cos^4 t}{(1+4\sin^2 t)^2}\) | 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) |
# Question 8:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 4\sec t \tan t$, $\quad \frac{dy}{dt} = 2\sec^2 t$ | B1 (both) | |
| $\frac{dy}{dx} = \frac{2\sec^2 t}{4\sec t \tan t} = \frac{1}{2\sin t}$ | M1 | |
| $y - 2\tan t = \frac{1}{2\sin t}(x - 4\sec t)$ | M1 A1 | |
| $2y\sin t - \frac{4\sin^2 t}{\cos t} = x - \frac{4}{\cos t}$ | | |
| $2y\sin t = x - \frac{4-4\sin^2 t}{\cos t} = x - 4\cos t$ ✱ | A1 | **(5 marks)** |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Gradient of $l_2$ is $-2\sin t$ | M1 | |
| $y = -2x\sin t$ | A1 | |
| $2(-2x\sin t)\sin t = x - 4\cos t \Rightarrow x = \frac{4\cos t}{1+4\sin^2 t}$ | M1 A1 | (1) |
| $y = \frac{-8\sin t \cos t}{1+4\sin^2 t}$ | M1 A1 | |
| $(x^2+y^2)^2 = \left(\frac{16\cos^2 t}{(1+4\sin^2 t)^2} + \frac{64\sin^2 t\cos^2 t}{(1+4\sin^2 t)^2}\right)^2$ | | |
| $= \frac{256\cos^4 t(1+4\sin^2 t)^2}{(1+4\sin^2 t)^4} = \frac{256\cos^4 t}{(1+4\sin^2 t)^2}$ | 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)** |\begin{enumerate}
\item The hyperbola $H$ has equation $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 1$.
\end{enumerate}
The line $l _ { 1 }$ is the tangent to $H$ at the point $P ( 4 \sec t , 2 \tan t )$.\\
(a) 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 $Q$.\\
(b) Show that, as $t$ varies, an equation of the locus of $Q$ is
$$\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 16 x ^ { 2 } - 4 y ^ { 2 }$$
\hfill \mbox{\textit{Edexcel F3 Q8 [13]}}