| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse tangent/normal equation derivation |
| Difficulty | Challenging +1.2 This is a standard Further Maths ellipse question requiring implicit differentiation to find the normal equation (part a) and then algebraic manipulation with eccentricity (part b). While it involves multiple steps and FP3 content, the techniques are routine for this level—implicit differentiation, perpendicular gradients, and eccentricity formulas are all standard procedures. The parametric form is given, making it more straightforward than if students had to derive it themselves. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2x}{36} + \frac{2y}{25}\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{25x}{36y} = \frac{5\cos\theta}{-6\sin\theta}\) | M1 | Correct attempt at \(\frac{dy}{dx}\) using implicit, parametric or explicit differentiation |
| \(= -\frac{5\cos\theta}{6\sin\theta}\) | A1 | Correct tangent gradient in terms of \(\theta\). May be implied by normal gradient |
| \(m_N = \frac{6\sin\theta}{5\cos\theta}\) | M1 | Correct perpendicular gradient rule. May be awarded if working in terms of \(x\) and \(y\) |
| \(y - 5\sin\theta = m_N(x - 6\cos\theta)\) | M1 | Correct straight line method for normal using a "changed" \(\frac{dy}{dx}\) in terms of \(\theta\) from calculus. If using \(y=mx+c\), must reach \(c=\ldots\) |
| \(6x\sin\theta - 5y\cos\theta = 11\sin\theta\cos\theta\) * | A1* | Correct completion to printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2 = a^2(1-e^2) \Rightarrow 25 = 36(1-e^2) \Rightarrow e^2 = \frac{11}{36}\), or \(e = \sqrt{\frac{11}{36}}\) | M1 | Uses the correct eccentricity formula to obtain a value for \(e\) or \(e^2\). Ignore \(\pm\) values for \(e\) |
| \(y=0 \Rightarrow x = \frac{11\cos\theta}{6}\) or \(\frac{11\sin\theta\cos\theta}{6\sin\theta}\) | B1 | Correct \(x\) coordinate for \(Q\) |
| \(\frac{OQ}{OR} = \frac{11\cos\theta}{6} \times \frac{1}{6\cos\theta}\) | M1 | Attempts \(\frac{\text{their }OQ}{\text{their }OR}\). May be implied by their ratio |
| \(= \frac{11}{36}\) | A1 | Correct completion with no errors to obtain \(\frac{11}{36}\) both times |
## Question 2(a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2x}{36} + \frac{2y}{25}\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{25x}{36y} = \frac{5\cos\theta}{-6\sin\theta}$ | M1 | Correct attempt at $\frac{dy}{dx}$ using implicit, parametric or explicit differentiation |
| $= -\frac{5\cos\theta}{6\sin\theta}$ | A1 | Correct tangent gradient in terms of $\theta$. May be implied by normal gradient |
| $m_N = \frac{6\sin\theta}{5\cos\theta}$ | M1 | Correct perpendicular gradient rule. May be awarded if working in terms of $x$ and $y$ |
| $y - 5\sin\theta = m_N(x - 6\cos\theta)$ | M1 | Correct straight line method for normal using a "changed" $\frac{dy}{dx}$ in terms of $\theta$ from calculus. If using $y=mx+c$, must reach $c=\ldots$ |
| $6x\sin\theta - 5y\cos\theta = 11\sin\theta\cos\theta$ * | A1* | Correct completion to printed answer with no errors |
**Note:** If candidate uses e.g. $y - 5\sin\theta = -\frac{36y}{25x}(x-6\cos\theta)$ before introducing $\theta$, the final mark can be withheld.
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## Question 2(b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2 = a^2(1-e^2) \Rightarrow 25 = 36(1-e^2) \Rightarrow e^2 = \frac{11}{36}$, or $e = \sqrt{\frac{11}{36}}$ | M1 | Uses the correct eccentricity formula to obtain a value for $e$ or $e^2$. Ignore $\pm$ values for $e$ |
| $y=0 \Rightarrow x = \frac{11\cos\theta}{6}$ or $\frac{11\sin\theta\cos\theta}{6\sin\theta}$ | B1 | Correct $x$ coordinate for $Q$ |
| $\frac{OQ}{OR} = \frac{11\cos\theta}{6} \times \frac{1}{6\cos\theta}$ | M1 | Attempts $\frac{\text{their }OQ}{\text{their }OR}$. May be implied by their ratio |
| $= \frac{11}{36}$ | A1 | Correct completion with no errors to obtain $\frac{11}{36}$ both times |
**Note:** Ignore any references to foci or directrices but the final mark can be withheld if there are any incorrect statements such as e.g. using $\cos\theta = 1$ in their ratio.
---2. The ellipse $E$ has equation
$$\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 25 } = 1$$
The line $l$ is the normal to $E$ at the point $P ( 6 \cos \theta , 5 \sin \theta )$, where $0 < \theta < \frac { \pi } { 2 }$
\begin{enumerate}[label=(\alph*)]
\item Use calculus to show that an equation of $l$ is
$$6 x \sin \theta - 5 y \cos \theta = 11 \sin \theta \cos \theta$$
The line $l$ meets the $x$-axis at the point $Q$.
The point $R$ is the foot of the perpendicular from $P$ to the $x$-axis.
\item Show that $\frac { O Q } { O R } = e ^ { 2 }$, where $e$ is the eccentricity of the ellipse $E$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2017 Q2 [9]}}