| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola tangent/normal equation derivation |
| Difficulty | Challenging +1.8 This is a Further Maths FP3 conic sections question requiring parametric differentiation, tangent equation derivation, and coordinate geometry proof. While the techniques are standard for FP3 (implicit/parametric differentiation, finding intersections, midpoint verification), it requires careful algebraic manipulation across multiple steps and coordination of trigonometric identities. The proof element in part (b) elevates it above routine exercises, but the path is relatively clear for well-prepared FM students. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07a Derivative as gradient: of tangent to curve1.07m Tangents and normals: gradient and equations |
| Answer | Marks |
|---|---|
| \(\frac{dx}{du} = 5\sec u \tan u\), \(\frac{dy}{du} = 3\sec^2 u\) | |
| \(\frac{dy}{dx} = \frac{3\sec^2 u}{5\sec u \tan u} = \frac{3}{5\sin u}\) | B1 |
| \(y - 3\tan u = \frac{3}{5\sin u}(x - 5\sec u)\) | M1 |
| Leading to \(3x = 5y\sin u + 15(\sec u - \tan u \sin u)\) | |
| \(3x = 5y\sin u + 15\left(\frac{1-\sin^2 u}{\cos u}\right)\) | |
| \(3x = 5y\sin u + 15\cos u\) ⭐ | M1 A1 cso (4) |
| Answer | Marks |
|---|---|
| Equations of asymptotes: \(y = \pm\frac{3}{5}x\) both | B1 |
| Answer | Marks |
|---|---|
| \(x = \frac{5\cos u}{1-\sin u}\), \(y = \frac{3\cos u}{1-\sin u}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(x = \frac{5\cos u}{1+\sin u}\), \(y = -\frac{3\cos u}{1+\sin u}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(x_M = \frac{1}{2}\left(\frac{5\cos u}{1-\sin u} + \frac{5\cos u}{1+\sin u}\right) = \frac{5\cos u}{2} \cdot \frac{2}{1-\sin^2 u} = 5\sec u\) | M1 |
| \(y_M = \frac{1}{2}\left(\frac{3\cos u}{1-\sin u} + \frac{-3\cos u}{1+\sin u}\right) = \frac{3\cos u}{2} \cdot \frac{2\sin u}{1-\sin^2 u} = 3\tan u\) | A1 |
| Answer | Marks |
|---|---|
| \(P\) is the mid-point of \(RS\). ⭐ | A1 cso (8) |
**(a)**
$\frac{dx}{du} = 5\sec u \tan u$, $\frac{dy}{du} = 3\sec^2 u$ | |
$\frac{dy}{dx} = \frac{3\sec^2 u}{5\sec u \tan u} = \frac{3}{5\sin u}$ | B1 |
$y - 3\tan u = \frac{3}{5\sin u}(x - 5\sec u)$ | M1 |
Leading to $3x = 5y\sin u + 15(\sec u - \tan u \sin u)$ | |
$3x = 5y\sin u + 15\left(\frac{1-\sin^2 u}{\cos u}\right)$ | |
$3x = 5y\sin u + 15\cos u$ ⭐ | M1 A1 cso (4) |
**(b)**
Equations of asymptotes: $y = \pm\frac{3}{5}x$ both | B1 |
Eliminating $y$ or $x$ between $3x = 5y\sin u + 15\cos u$ and $y = \frac{3}{5}x$:
$3x = 3x\sin u + 15\cos u$
$x = \frac{5\cos u}{1-\sin u}$, $y = \frac{3\cos u}{1-\sin u}$ | M1 A1 |
Similarly between $3x = 5y\sin u + 15\cos u$ and $y = -\frac{3}{5}x$:
$x = \frac{5\cos u}{1+\sin u}$, $y = -\frac{3\cos u}{1+\sin u}$ | M1 A1 |
Let $(x_M, y_M)$ be the coordinates of the mid-point of $RS$.
$x_M = \frac{1}{2}\left(\frac{5\cos u}{1-\sin u} + \frac{5\cos u}{1+\sin u}\right) = \frac{5\cos u}{2} \cdot \frac{2}{1-\sin^2 u} = 5\sec u$ | M1 |
$y_M = \frac{1}{2}\left(\frac{3\cos u}{1-\sin u} + \frac{-3\cos u}{1+\sin u}\right) = \frac{3\cos u}{2} \cdot \frac{2\sin u}{1-\sin^2 u} = 3\tan u$ | A1 |
The coordinates $(x_M, y_M)$ are the same as $P$.
$P$ is the mid-point of $RS$. ⭐ | A1 cso (8) |
(12 marks)8. The point $\mathrm { P } ( 5 \sec \mathrm { u } , 3 \tan \mathrm { u } )$ lies on the hyperbola H with equation $\frac { \mathrm { x } ^ { 2 } } { 25 } - \frac { \mathrm { y } ^ { 2 } } { 9 } = 1$.
The tangent to $H$ at $P$ intersects the asymptote of $H$ with equation $y = \frac { 3 } { 5 } x$ at the point $R$ and the asymptote with equation $\mathrm { y } = - \frac { 3 } { 5 } \mathrm { x }$ at the point S .
\begin{enumerate}[label=(\alph*)]
\item Use differentiation to show that an equation of the tangent to H at P is
$$3 x = 5 y \sin u + 15 \cos u$$
\item Prove that P is the mid-point of RS.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 Q8 [12]}}