| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola locus problems |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring implicit differentiation of a hyperbola, parametric normal equation derivation, coordinate geometry to find axis intercepts, and locus determination. While the techniques are standard for FP3, the multi-step nature, algebraic manipulation demands, and locus work make it significantly harder than typical A-level questions but not exceptionally difficult for Further Maths students. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) | ||
| \(\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0\) | M1A1 | |
| \(\frac{dy}{dx} = \frac{2x b^2}{a^2 2y} = \frac{b^2}{a^2}\operatorname{asec}\theta = \frac{b}{a\sin\theta}\) | M1A1 | |
| Gradient of normal is then \(-\frac{a}{b}\sin\theta\) | ||
| Equation of normal: \((y - b\tan\theta) = -\frac{a}{b}\sin\theta(x - a\sec\theta)\) | ||
| \(ax\sin\theta + by = (a^2 + b^2)\tan\theta\) | M1A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| M: A normal cuts \(x = 0\) at \(y = \frac{(a^2 + b^2)}{b}\tan\theta\) | M1A1 | |
| B normal cuts \(y = 0\) at \(x = \frac{a^2 + b^2}{a\sin\theta}\tan\theta = \frac{(a^2 + b^2)}{a\cos\theta}\) | A1 | |
| Hence M is \(\left[\frac{(a^2 + b^2)}{2a}\sec\theta, \frac{(a^2 + b^2)}{2b}\tan\theta\right]\) | M1 | |
| Eliminating \(\theta\): \(\sec^2\theta = 1 + \tan^2\theta\) | M1 | |
| \(\left[\frac{2aX}{a^2 + b^2}\right]^2 = 1 + \left[\frac{2bY}{a^2 + b^2}\right]^2\) | A1 | |
| \(4a^2X^2 - 4b^2Y^2 = [a^2 + b^2]^2\) | A1 | (7) |
## (a)
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ | |
$\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} = 0$ | M1A1 |
$\frac{dy}{dx} = \frac{2x b^2}{a^2 2y} = \frac{b^2}{a^2}\operatorname{asec}\theta = \frac{b}{a\sin\theta}$ | M1A1 |
Gradient of normal is then $-\frac{a}{b}\sin\theta$ | |
Equation of normal: $(y - b\tan\theta) = -\frac{a}{b}\sin\theta(x - a\sec\theta)$ | |
$ax\sin\theta + by = (a^2 + b^2)\tan\theta$ | M1A1 | (6)
## (b)
M: A normal cuts $x = 0$ at $y = \frac{(a^2 + b^2)}{b}\tan\theta$ | M1A1 |
B normal cuts $y = 0$ at $x = \frac{a^2 + b^2}{a\sin\theta}\tan\theta = \frac{(a^2 + b^2)}{a\cos\theta}$ | A1 |
Hence M is $\left[\frac{(a^2 + b^2)}{2a}\sec\theta, \frac{(a^2 + b^2)}{2b}\tan\theta\right]$ | M1 |
Eliminating $\theta$: $\sec^2\theta = 1 + \tan^2\theta$ | M1 |
$\left[\frac{2aX}{a^2 + b^2}\right]^2 = 1 + \left[\frac{2bY}{a^2 + b^2}\right]^2$ | A1 |
$4a^2X^2 - 4b^2Y^2 = [a^2 + b^2]^2$ | A1 | (7)The hyperbola $C$ has equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the normal to $C$ at $P(a \sec \theta, b \tan \theta)$ is
$$by + ax \sin \theta = (a^2 + b^2)\tan \theta$$ [6]
The normal at $P$ cuts the coordinate axes at $A$ and $B$. The mid-point of $AB$ is $M$.
\item Find, in cartesian form, an equation of the locus of $M$ as $\theta$ varies. [7]
\end{enumerate}
(Total 13 marks)
\hfill \mbox{\textit{Edexcel FP3 Q9 [13]}}