| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse tangent/normal equation derivation |
| Difficulty | Challenging +1.2 This is a Further Maths question on ellipse normals using parametric form, requiring implicit differentiation, normal equation derivation, and coordinate geometry for area calculation. Part (a) is a standard 'show that' requiring routine techniques, while part (b) involves finding coordinates and computing a triangle area. The multi-step nature and Further Maths context place it above average difficulty, but the techniques are well-practiced for FM students with no novel insight required. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03e Complete the square: find centre and radius of circle1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(\frac{dx}{d\theta} = -5\sin\theta\), \(\frac{dy}{d\theta} = 2\cos\theta\) or \(\frac{2x}{25} + \frac{2y}{4}\frac{dy}{dx} = 0\) | B1 | Correct derivatives or correct implicit differentiation |
| \(\frac{dy}{dx} = \frac{2\cos\theta}{-5\sin\theta}\) | M1 | Divides their derivatives correctly or substitutes and rearranges |
| \(M_N = \frac{5\sin\theta}{2\cos\theta}\) | M1 | Correct perpendicular gradient rule |
| \(y - 2\sin\theta = \frac{5\sin\theta}{2\cos\theta}(x - 5\cos\theta)\) | M1 | Correct straight line method (any complete method); Must use their gradient of the normal |
| \(5x\sin\theta - 2y\cos\theta = 21\sin\theta\cos\theta\)* | A1* | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| At \(Q\), \(x = 0 \Rightarrow y = -\frac{21}{2}\sin\theta\) | B1 | |
| \(M\) is \(\left(\frac{0 + 5\cos\theta}{2}, \frac{2\sin\theta - \frac{21}{2}\sin\theta}{2}\right) = \left(\frac{5}{2}\cos\theta, -\frac{17}{4}\sin\theta\right)\) | M1 | Correct mid-point method for at least one coordinate; can be implied by correct \(x\) coordinate |
| \(L\) cuts \(x\)-axis at \(\frac{21}{5}\cos\theta\) | B1 | |
| Area \(OPM =\) Area \(OLP +\) Area \(OLM\) | M1A1 | M1: Correct triangle area method using their coordinates; A1: Correct expression |
| \(\frac{1}{2}\cdot\frac{21}{5}\cos\theta\cdot 2\sin\theta + \frac{1}{2}\cdot\frac{21}{5}\cos\theta\cdot\frac{17}{4}\sin\theta\) | ||
| \(= \frac{105}{16}\sin 2\theta\) | A1 | Or \(6.5625\sin 2\theta\); must be positive |
## Question 2:
$\frac{x^2}{25} + \frac{y^2}{4} = 1$, $P(5\cos\theta, 2\sin\theta)$
### Part (a):
| Answer/Working | Mark | Notes |
|---|---|---|
| $\frac{dx}{d\theta} = -5\sin\theta$, $\frac{dy}{d\theta} = 2\cos\theta$ or $\frac{2x}{25} + \frac{2y}{4}\frac{dy}{dx} = 0$ | B1 | Correct derivatives or correct implicit differentiation |
| $\frac{dy}{dx} = \frac{2\cos\theta}{-5\sin\theta}$ | M1 | Divides their derivatives correctly or substitutes and rearranges |
| $M_N = \frac{5\sin\theta}{2\cos\theta}$ | M1 | Correct perpendicular gradient rule |
| $y - 2\sin\theta = \frac{5\sin\theta}{2\cos\theta}(x - 5\cos\theta)$ | M1 | Correct straight line method (any complete method); **Must** use their gradient of the normal |
| $5x\sin\theta - 2y\cos\theta = 21\sin\theta\cos\theta$* | A1* | cso |
### Part (b):
| Answer/Working | Mark | Notes |
|---|---|---|
| At $Q$, $x = 0 \Rightarrow y = -\frac{21}{2}\sin\theta$ | B1 | |
| $M$ is $\left(\frac{0 + 5\cos\theta}{2}, \frac{2\sin\theta - \frac{21}{2}\sin\theta}{2}\right) = \left(\frac{5}{2}\cos\theta, -\frac{17}{4}\sin\theta\right)$ | M1 | Correct mid-point method for at least one coordinate; can be implied by correct $x$ coordinate |
| $L$ cuts $x$-axis at $\frac{21}{5}\cos\theta$ | B1 | |
| Area $OPM =$ Area $OLP +$ Area $OLM$ | M1A1 | M1: Correct triangle area method using their coordinates; A1: Correct expression |
| $\frac{1}{2}\cdot\frac{21}{5}\cos\theta\cdot 2\sin\theta + \frac{1}{2}\cdot\frac{21}{5}\cos\theta\cdot\frac{17}{4}\sin\theta$ | | |
| $= \frac{105}{16}\sin 2\theta$ | A1 | Or $6.5625\sin 2\theta$; must be positive |
---2. An ellipse has equation
$$\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 4 } = 1$$
The point $P$ lies on the ellipse and has coordinates $( 5 \cos \theta , 2 \sin \theta ) , 0 < \theta < \frac { \pi } { 2 }$ The line $L$ is a normal to the ellipse at the point $P$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation for $L$ is
$$5 x \sin \theta - 2 y \cos \theta = 21 \sin \theta \cos \theta$$
Given that the line $L$ crosses the $y$-axis at the point $Q$ and that $M$ is the midpoint of $P Q$,
\item find the exact area of triangle $O P M$, where $O$ is the origin, giving your answer as a multiple of $\sin 2 \theta$\\
WIHN SIHI NITIIUM ION OC\\
VIUV SIHI NI JAHM ION OC\\
VI4V SIHI NIS IIIM ION OC
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2016 Q2 [11]}}