| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Hyperbola focus-directrix properties |
| Difficulty | Challenging +1.8 This is a substantial Further Maths conic sections question requiring multiple techniques: standard hyperbola properties (parts a-b), parametric differentiation to find tangent equations (part c), and coordinate geometry with simultaneous equations to derive a locus (part d). The final part requires algebraic manipulation across 7 marks to eliminate the parameter and identify the ellipse, which is non-trivial but follows a clear method for FM students familiar with conics. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | Asymptotes are y =±2x | y |
| Answer | Marks |
|---|---|
| 2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 8(b) | 4 = e2 −1⇒ e = 5 | Uses the correct eccentricity formula with a |
| = 1 and b = 2 to find a value for e. | M1 | |
| Foci are (± 5,0) | Both required. | A1 |
| Answer | Marks |
|---|---|
| 8(c) | dy dy 4x 4secθ dy dy dθ 2sec2θ |
| Answer | Marks |
|---|---|
| A1: Correct gradient in terms of θ | M1A1 |
| Answer | Marks |
|---|---|
| 2tanθ | Correct straight line method using their |
| Answer | Marks |
|---|---|
| y = 2tan θ | M1 |
| Answer | Marks |
|---|---|
| ytanθ= 2xsecθ−2* | Obtains the given answer with sufficient |
| working shown as above. | A1cso |
| Answer | Marks |
|---|---|
| 8(d) | 2tanθ |
| Answer | Marks |
|---|---|
| A1: One correct equation in any form | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| secθ−1 secθ+1 | Both equations correct in any form. | A1 |
| Answer | Marks |
|---|---|
| secθ+1 secθ−1 | Attempt to solve and makes progress to |
| Answer | Marks | Guidance |
|---|---|---|
| only. | M1 | |
| x=cosθ or y=2sinθ | One correct coordinate | A1 |
| x=cosθ and y=2sinθ | Both correct | A1 |
| Answer | Marks |
|---|---|
| 4 | Correct equation or correct values for a and |
| b | A1 |
Total 14
Question 8:
--- 8(a) ---
8(a) | Asymptotes are y =±2x | y
y =±2xoe e.g. x = ±
2 | B1
(1)
--- 8(b) ---
8(b) | 4 = e2 −1⇒ e = 5 | Uses the correct eccentricity formula with a
= 1 and b = 2 to find a value for e. | M1
Foci are (± 5,0) | Both required. | A1
(2)
--- 8(c) ---
8(c) | dy dy 4x 4secθ dy dy dθ 2sec2θ
8x−2y =0⇒ = = or = × =
dx dx y 2tanθ dx dθ dx secθtanθ
M1:Ax+By dy =0⇒ dy =f(θ) or dy = dy × dθ =f(θ)
dx dx dx dθ dx
A1: Correct gradient in terms of θ | M1A1
Explicit differentiation may be seen:
y2 =4x2−4⇒ y= ( 4x2−4 ) 1 2 ⇒ dy = 1( 4x2−4 )− 1 2×8x= 4secθ
dx 2 4sec2θ−4
1
Score M1 for dy = kx ( 4x2 −4 )− 2 = f (θ) and A1 for correct gradient in terms of θ
dx
4secθ
E.g. y−2tanθ= (x−secθ)
2tanθ | Correct straight line method using their
gradient in terms of θ and x = sec θ,
y = 2tan θ | M1
ytanθ−2tan2θ=2xsecθ−2sec2θ
⇒ ytanθ−2tan2θ=2xsecθ−2(1+tan2θ)
ytanθ= 2xsecθ−2* | Obtains the given answer with sufficient
working shown as above. | A1cso
(4)
--- 8(d) ---
8(d) | 2tanθ
VP:V(−1,0);P(secθ,2tanθ)⇒ y = (x+1)
secθ+1
or
−2tanθ
WQ:W(1,0);Q(secθ.−2tanθ)⇒ y = (x−1)
secθ−1
M1: Correct straight line method for either VP or WQ
A1: One correct equation in any form | M1A1
−2tanθ 2tanθ
y = (x−1), y = (x+1)
secθ−1 secθ+1 | Both equations correct in any form. | A1
2tanθ −2tanθ
(x+1)= (x−1)⇒ x/ y =...
secθ+1 secθ−1 | Attempt to solve and makes progress to
achieve either x = … or y = …in terms of θ
only. | M1
x=cosθ or y=2sinθ | One correct coordinate | A1
x=cosθ and y=2sinθ | Both correct | A1
y2
x2 + =1 or a =1,b =2
4 | Correct equation or correct values for a and
b | A1
(7)
Total 14
PMT
Pearson Education Limited. Registered company number 872828
with its registered office at 80 Strand, London, WC2R 0RL, United KingdomThe hyperbola $H$ has equation
$$4x^2 - y^2 = 4$$
\begin{enumerate}[label=(\alph*)]
\item Write down the equations of the asymptotes of $H$.
[1]
\item Find the coordinates of the foci of $H$.
[2]
\end{enumerate}
The point $P(\sec \theta, 2 \tan \theta)$ lies on $H$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Using calculus, show that the equation of the tangent to $H$ at the point $P$ is
$$y \tan \theta = 2x \sec \theta - 2$$
[4]
\end{enumerate}
The point $V(-1, 0)$ and the point $W(1, 0)$ both lie on $H$.
The point $Q(\sec \theta, -2 \tan \theta)$ also lies on $H$.
Given that $P$, $Q$, $V$ and $W$ are distinct points on $H$ and that the lines $VP$ and $WQ$ intersect at the point $S$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item show that, as $\theta$ varies, $S$ lies on an ellipse with equation
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
where $a$ and $b$ are integers to be found.
[7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2021 Q8 [14]}}