| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse tangent/normal equation derivation |
| Difficulty | Challenging +1.2 This is a structured Further Maths ellipse question requiring parametric differentiation for the normal equation, standard focus formula application, and distance ratio manipulation using eccentricity. While it involves multiple parts and Further Maths content (making it harder than average A-level), the techniques are standard for F3 with clear scaffolding through parts (a), (b), and (c). The 'show that' format provides targets to work towards, reducing problem-solving demand. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.03h Parametric equations: in modelling contexts1.07s Parametric and implicit differentiation |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{d\theta}=-5\sin\theta,\quad\frac{dy}{d\theta}=4\cos\theta\) or \(\frac{2x}{25}+\frac{2y}{16}\frac{dy}{dx}=0\) or \(\frac{dy}{dx}=-\frac{4x}{25}\left(1-\frac{x^2}{25}\right)^{-\frac{1}{2}}\) | B1 | Correct derivatives or correct implicit/explicit differentiation |
| \(\frac{dy}{dx}=\frac{4\cos\theta}{-5\sin\theta}\) | M1 | Divides their derivatives correctly or substitutes and rearranges |
| \(M_N = \frac{5\sin\theta}{4\cos\theta}\) | M1 | Correct perpendicular gradient rule; may be implied when forming normal equation |
| \(y-4\sin\theta = \frac{5\sin\theta}{4\cos\theta}(x-5\cos\theta)\) | M1 | Correct straight line method (any complete method); must use their gradient of the normal |
| \(5x\sin\theta - 4y\cos\theta = 9\sin\theta\cos\theta\) or \(9\sin\theta\cos\theta = 5x\sin\theta - 4y\cos\theta\) | A1* | Achieves printed answer with no errors; allow \(5\sin\theta x\) for \(5x\sin\theta\) and \(4\cos\theta y\) for \(4y\cos\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(b^2=a^2(1-e^2)\Rightarrow 16=25(1-e^2)\Rightarrow e=\frac{3}{5}\); \(F\) is \((ae,0)=\left(5\times\frac{3}{5},0\right)\) or \(c^2=a^2e^2=a^2-b^2=25-16=9\Rightarrow ae=\ldots\) | M1 | Fully correct strategy for \(F\) (must be numerical so \((5e,0)\) is M0) |
| \((3,0)\) | A1 | Correct coordinates; \((\pm3,0)\) scores A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=\frac{9}{5}\cos\theta\) | B1 | Correct \(x\) coordinate of \(Q\) |
| \(PF^2=(5\cos\theta-\text{"3"})^2+(4\sin\theta)^2\) or \(PF=\sqrt{(5\cos\theta-\text{"3"})^2+(4\sin\theta)^2}\) | M1 | Correct application of Pythagoras to find \(PF\) or \(PF^2\); "3" should be positive but allow work in terms of \(e\), e.g. "5e" |
| \(=25\cos^2\theta-30\cos\theta+9+16\sin^2\theta = 25\cos^2\theta-30\cos\theta+9+16(1-\cos^2\theta)\) | dM1 | Applies \(\sin^2\theta=1-\cos^2\theta\) to obtain quadratic in \(\cos\theta\); if identity not seen explicitly, working must imply correct identity used; depends on previous M |
| \(PF=\pm(5-3\cos\theta)\), \(PF^2=9\cos^2\theta-30\cos\theta+25\) | A1 | Correct expression for \(PF\) or \(PF^2\) in terms of \(\cos\theta\) with terms collected |
| \(\frac{\ | QF\ | }{\ |
## Question 9:
**Ellipse:** $\frac{x^2}{25}+\frac{y^2}{16}=1$, parametric form $(5\cos\theta, 4\sin\theta)$
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{d\theta}=-5\sin\theta,\quad\frac{dy}{d\theta}=4\cos\theta$ **or** $\frac{2x}{25}+\frac{2y}{16}\frac{dy}{dx}=0$ **or** $\frac{dy}{dx}=-\frac{4x}{25}\left(1-\frac{x^2}{25}\right)^{-\frac{1}{2}}$ | B1 | Correct derivatives or correct implicit/explicit differentiation |
| $\frac{dy}{dx}=\frac{4\cos\theta}{-5\sin\theta}$ | M1 | Divides their derivatives correctly or substitutes and rearranges |
| $M_N = \frac{5\sin\theta}{4\cos\theta}$ | M1 | Correct perpendicular gradient rule; may be implied when forming normal equation |
| $y-4\sin\theta = \frac{5\sin\theta}{4\cos\theta}(x-5\cos\theta)$ | M1 | Correct straight line method (any complete method); **must** use their gradient of the normal |
| $5x\sin\theta - 4y\cos\theta = 9\sin\theta\cos\theta$ **or** $9\sin\theta\cos\theta = 5x\sin\theta - 4y\cos\theta$ | A1* | Achieves printed answer with no errors; allow $5\sin\theta x$ for $5x\sin\theta$ and $4\cos\theta y$ for $4y\cos\theta$ |
**(5 marks)**
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b^2=a^2(1-e^2)\Rightarrow 16=25(1-e^2)\Rightarrow e=\frac{3}{5}$; $F$ is $(ae,0)=\left(5\times\frac{3}{5},0\right)$ **or** $c^2=a^2e^2=a^2-b^2=25-16=9\Rightarrow ae=\ldots$ | M1 | Fully correct strategy for $F$ (must be numerical so $(5e,0)$ is M0) |
| $(3,0)$ | A1 | Correct coordinates; $(\pm3,0)$ scores A0 |
**(2 marks)**
---
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\frac{9}{5}\cos\theta$ | B1 | Correct $x$ coordinate of $Q$ |
| $PF^2=(5\cos\theta-\text{"3"})^2+(4\sin\theta)^2$ **or** $PF=\sqrt{(5\cos\theta-\text{"3"})^2+(4\sin\theta)^2}$ | M1 | Correct application of Pythagoras to find $PF$ or $PF^2$; "3" should be positive but allow work in terms of $e$, e.g. "5e" |
| $=25\cos^2\theta-30\cos\theta+9+16\sin^2\theta = 25\cos^2\theta-30\cos\theta+9+16(1-\cos^2\theta)$ | dM1 | Applies $\sin^2\theta=1-\cos^2\theta$ to obtain quadratic in $\cos\theta$; if identity not seen explicitly, working must imply correct identity used; **depends on previous M** |
| $PF=\pm(5-3\cos\theta)$, $PF^2=9\cos^2\theta-30\cos\theta+25$ | A1 | Correct expression for $PF$ or $PF^2$ in terms of $\cos\theta$ with terms collected |
| $\frac{\|QF\|}{\|PF\|}=\frac{3-\frac{9}{5}\cos\theta}{5-3\cos\theta}=\frac{3\left(1-\frac{3}{5}\cos\theta\right)}{5\left(1-\frac{3}{5}\cos\theta\right)}=\frac{3}{5}=e$ | A1* | Fully correct working including factorisation leading to $\frac{\|QF\|}{\|PF\|}=e$ with no errors and conclusion "$=e$" |
**Alternative using** $PF=ePM$:
Score M1 for $x=\frac{a}{e}=\frac{5}{\frac{3}{5}}=\frac{25}{3}$ (not $\pm\frac{25}{3}$) and dM1A1 for $PF=ePM=\frac{3}{5}\left(\frac{25}{3}-5\cos\theta\right)$
**(5 marks)**
**Total: 12 marks**9. The ellipse $E$ has equation
$$\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 16 } = 1$$
The point $P$ lies on the ellipse and has coordinates $( 5 \cos \theta , 4 \sin \theta )$ where $0 < \theta < \frac { \pi } { 2 }$
The line $l$ is the normal to the ellipse at the point $P$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation for $l$ is
$$5 x \sin \theta - 4 y \cos \theta = 9 \sin \theta \cos \theta$$
The point $F$ is the focus of $E$ that lies on the positive $x$-axis.
\item Determine the coordinates of $F$.
The line $l$ crosses the $x$-axis at the point $Q$.
\item Show that
$$\frac { | Q F | } { | P F | } = e$$
where $e$ is the eccentricity of $E$.\\
\begin{center}
\begin{tabular}{|l|l|}
\hline
\hline
END & \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2021 Q9 [12]}}