| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse tangent/normal equation derivation |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring parametric ellipse representation, implicit differentiation for tangent equations, coordinate geometry for intercepts and area, and locus derivation. While systematic, it demands fluency with multiple techniques and algebraic manipulation across four connected parts, placing it moderately above average difficulty. |
| Spec | 1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a=3,\quad b=1\) | B1 | Both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = -\dfrac{\cos\theta}{3\sin\theta}\) | M1 | Complete correct gradient method including use of coordinates |
| \(y - \sin\theta = -\dfrac{\cos\theta}{3\sin\theta}(x-3\cos\theta)\quad\text{(I)}\) | M1 | Correct straight line method |
| \(3y\sin\theta - 3\sin^2\theta = -x\cos\theta + 3\cos^2\theta\) | Allow both M's if working in \(a\) and \(b\) so far | |
| \(3y\sin\theta + x\cos\theta = 3\cos^2\theta + 3\sin^2\theta = 3\;*\) | A1* | Correct completion to printed answer with no errors seen; some working needed from (I) to * |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=0 \Rightarrow y=\dfrac{1}{\sin\theta},\quad y=0 \Rightarrow x=\dfrac{3}{\cos\theta}\) | B1 | Both |
| \(\text{Area}=\dfrac{1}{2}\times\dfrac{1}{\sin\theta}\times\dfrac{3}{\cos\theta}\) | M1 | Correct method for area |
| \(= \dfrac{3}{2\sin\theta\cos\theta} = 3\text{cosec}\,2\theta\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=\dfrac{3}{2\cos\theta},\quad y=\dfrac{1}{2\sin\theta}\) | B1ft | Correct follow-through mid-point |
| \(\sin\theta=\dfrac{1}{2y},\quad \cos\theta=\dfrac{3}{2x}\); then \(\left(\dfrac{3}{2x}\right)^2+\left(\dfrac{1}{2y}\right)^2=1\) | M1 | Attempt \(\sin\) and \(\cos\) in terms of \(x\) and \(y\) and attempt Pythagoras; allow if \(x\) and \(y\) are exchanged |
| \(9y^2+x^2=4x^2y^2\) | ||
| \(y^2(4x^2-9)=x^2 \Rightarrow y^2=\ldots\) | M1 | Attempt to isolate \(y^2\) |
| \(y^2=\dfrac{x^2}{4x^2-9}\) | A1 | Correct equation (oe) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=\dfrac{3}{2\cos\theta},\quad y=\dfrac{1}{2\sin\theta}\) | B1ft | Correct follow-through mid-point |
| \(y^2=\dfrac{1}{4\sin^2\theta}\) | M1 | Attempt \(y^2\) in terms of \(\sin\) |
| \(y^2=\dfrac{1}{4(1-\cos^2\theta)}\) | M1 | Correct use of Pythagoras |
| \(y^2=\dfrac{1}{4\!\left(1-\dfrac{9}{4x^2}\right)}\) | A1 | Correct equation (oe) |
# Question 5
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a=3,\quad b=1$ | B1 | Both |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = -\dfrac{\cos\theta}{3\sin\theta}$ | M1 | Complete correct gradient method including use of coordinates |
| $y - \sin\theta = -\dfrac{\cos\theta}{3\sin\theta}(x-3\cos\theta)\quad\text{(I)}$ | M1 | Correct straight line method |
| $3y\sin\theta - 3\sin^2\theta = -x\cos\theta + 3\cos^2\theta$ | | Allow both M's if working in $a$ and $b$ so far |
| $3y\sin\theta + x\cos\theta = 3\cos^2\theta + 3\sin^2\theta = 3\;*$ | A1* | Correct completion to printed answer with no errors seen; some working needed from (I) to * |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=0 \Rightarrow y=\dfrac{1}{\sin\theta},\quad y=0 \Rightarrow x=\dfrac{3}{\cos\theta}$ | B1 | Both |
| $\text{Area}=\dfrac{1}{2}\times\dfrac{1}{\sin\theta}\times\dfrac{3}{\cos\theta}$ | M1 | Correct method for area |
| $= \dfrac{3}{2\sin\theta\cos\theta} = 3\text{cosec}\,2\theta$ | A1 | |
**Part (d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\dfrac{3}{2\cos\theta},\quad y=\dfrac{1}{2\sin\theta}$ | B1ft | Correct follow-through mid-point |
| $\sin\theta=\dfrac{1}{2y},\quad \cos\theta=\dfrac{3}{2x}$; then $\left(\dfrac{3}{2x}\right)^2+\left(\dfrac{1}{2y}\right)^2=1$ | M1 | Attempt $\sin$ and $\cos$ in terms of $x$ and $y$ and attempt Pythagoras; allow if $x$ and $y$ are exchanged |
| $9y^2+x^2=4x^2y^2$ | | |
| $y^2(4x^2-9)=x^2 \Rightarrow y^2=\ldots$ | M1 | Attempt to isolate $y^2$ |
| $y^2=\dfrac{x^2}{4x^2-9}$ | A1 | Correct equation (oe) |
**Part (d) Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\dfrac{3}{2\cos\theta},\quad y=\dfrac{1}{2\sin\theta}$ | B1ft | Correct follow-through mid-point |
| $y^2=\dfrac{1}{4\sin^2\theta}$ | M1 | Attempt $y^2$ in terms of $\sin$ |
| $y^2=\dfrac{1}{4(1-\cos^2\theta)}$ | M1 | Correct use of Pythagoras |
| $y^2=\dfrac{1}{4\!\left(1-\dfrac{9}{4x^2}\right)}$ | A1 | Correct equation (oe) |5. The ellipse $E$ has equation
$$x ^ { 2 } + 9 y ^ { 2 } = 9$$
The point $P ( a \cos \theta , b \sin \theta )$ is a general point on the ellipse $E$.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $a$ and the value of $b$.
The line $L$ is a tangent to $E$ at the point $P$.
\item Show that an equation of the line $L$ is given by
$$3 y \sin \theta + x \cos \theta = 3$$
The line $L$ meets the $x$-axis at the point $Q$ and meets the $y$-axis at the point $R$.
\item Show that the area of the triangle $O Q R$, where $O$ is the origin, is given by
$$k \operatorname { cosec } 2 \theta$$
where $k$ is a constant to be found.
The point $M$ is the midpoint of $Q R$.
\item Find a cartesian equation of the locus of $M$, giving your answer in the form $y ^ { 2 } = \mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2014 Q5 [11]}}